The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

Question

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Answer 1

Question

Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

(a)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $90 ml$ and of barium chloride solution is $500 ml$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $90 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$90 mL=1 L1000 mL×90 mL=0.09 L$

Convert $500 ml$ volume of barium chloride solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Molarity of the solution $M$ is equal to the ratio of the number of moles $n$ of a solute to the volume $V$ of the solution in litres.

$M=nV$ â€¦â€¦ (1)

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.5 Ln=0.1 mol L−1×0.5 L=0.05 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.09 L$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$0.5 mol L−1=n0.09 Ln=0.5 mol L−1×0.09 L=0.045 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4$

Barium sulfate formed is as follows:

$1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4$

However, only $0.045 mol$ of $K2SO4$ is present. Therefore, $0.045 mol$ of $K2SO4$ reacts with $0.045 mol$ of $BaCl2$ to produce $0.045 mol$ of $BaSO4$ .

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

The number of moles $n$ of a substance is equal to the ratio of the mass $m$ of that substance to the molar mass $MM$ of the substance.

$n=mMM$ â€¦â€¦ (2)

Substitute $n$ as $0.045 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.045 mol=m233.38 g mol−1m=0.045 mol×233.38 g mol−1=10.5021 g$

Thus, the mass of barium sulfate formed is $l0.502l g$ .

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(b)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $100 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $100 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.01 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.1 Ln=0.5 mol L−1×0.1 L=0.05 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The number of moles of $BaSO4$ produced as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(c)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $500 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $500 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.5 Ln=0.5 mol L−1×0.5 L=0.25 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$∴ 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2. ∴ 0.01 moles of BaCl2 reacts with 0.01 moles of K2 SO4 to form 0.01 moles of BaSO4$

$1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2$

However, the number of moles of $BaCl2$ is only $0.01 mol$ .

$0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The amount of $BaSO4$ formed is as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

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Answer 2

Question

Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

(a)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $90 ml$ and of barium chloride solution is $500 ml$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $90 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$90 mL=1 L1000 mL×90 mL=0.09 L$

Convert $500 ml$ volume of barium chloride solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Molarity of the solution $M$ is equal to the ratio of the number of moles $n$ of a solute to the volume $V$ of the solution in litres.

$M=nV$ â€¦â€¦ (1)

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.5 Ln=0.1 mol L−1×0.5 L=0.05 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.09 L$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$0.5 mol L−1=n0.09 Ln=0.5 mol L−1×0.09 L=0.045 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4$

Barium sulfate formed is as follows:

$1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4$

However, only $0.045 mol$ of $K2SO4$ is present. Therefore, $0.045 mol$ of $K2SO4$ reacts with $0.045 mol$ of $BaCl2$ to produce $0.045 mol$ of $BaSO4$ .

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

The number of moles $n$ of a substance is equal to the ratio of the mass $m$ of that substance to the molar mass $MM$ of the substance.

$n=mMM$ â€¦â€¦ (2)

Substitute $n$ as $0.045 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.045 mol=m233.38 g mol−1m=0.045 mol×233.38 g mol−1=10.5021 g$

Thus, the mass of barium sulfate formed is $l0.502l g$ .

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(b)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $100 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $100 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.01 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.1 Ln=0.5 mol L−1×0.1 L=0.05 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The number of moles of $BaSO4$ produced as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(c)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $500 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $500 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.5 Ln=0.5 mol L−1×0.5 L=0.25 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$∴ 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2. ∴ 0.01 moles of BaCl2 reacts with 0.01 moles of K2 SO4 to form 0.01 moles of BaSO4$

$1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2$

However, the number of moles of $BaCl2$ is only $0.01 mol$ .

$0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The amount of $BaSO4$ formed is as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

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Answer 3

Question

Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

(a)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $90 ml$ and of barium chloride solution is $500 ml$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $90 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$90 mL=1 L1000 mL×90 mL=0.09 L$

Convert $500 ml$ volume of barium chloride solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Molarity of the solution $M$ is equal to the ratio of the number of moles $n$ of a solute to the volume $V$ of the solution in litres.

$M=nV$ â€¦â€¦ (1)

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.5 Ln=0.1 mol L−1×0.5 L=0.05 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.09 L$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$0.5 mol L−1=n0.09 Ln=0.5 mol L−1×0.09 L=0.045 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4$

Barium sulfate formed is as follows:

$1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4$

However, only $0.045 mol$ of $K2SO4$ is present. Therefore, $0.045 mol$ of $K2SO4$ reacts with $0.045 mol$ of $BaCl2$ to produce $0.045 mol$ of $BaSO4$ .

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

The number of moles $n$ of a substance is equal to the ratio of the mass $m$ of that substance to the molar mass $MM$ of the substance.

$n=mMM$ â€¦â€¦ (2)

Substitute $n$ as $0.045 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.045 mol=m233.38 g mol−1m=0.045 mol×233.38 g mol−1=10.5021 g$

Thus, the mass of barium sulfate formed is $l0.502l g$ .

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(b)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $100 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $100 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.01 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.1 Ln=0.5 mol L−1×0.1 L=0.05 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The number of moles of $BaSO4$ produced as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(c)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $500 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $500 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.5 Ln=0.5 mol L−1×0.5 L=0.25 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$∴ 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2. ∴ 0.01 moles of BaCl2 reacts with 0.01 moles of K2 SO4 to form 0.01 moles of BaSO4$

$1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2$

However, the number of moles of $BaCl2$ is only $0.01 mol$ .

$0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The amount of $BaSO4$ formed is as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

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Answer 4

Question

Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

(a)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $90 ml$ and of barium chloride solution is $500 ml$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $90 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$90 mL=1 L1000 mL×90 mL=0.09 L$

Convert $500 ml$ volume of barium chloride solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Molarity of the solution $M$ is equal to the ratio of the number of moles $n$ of a solute to the volume $V$ of the solution in litres.

$M=nV$ â€¦â€¦ (1)

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.5 Ln=0.1 mol L−1×0.5 L=0.05 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.09 L$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$0.5 mol L−1=n0.09 Ln=0.5 mol L−1×0.09 L=0.045 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4$

Barium sulfate formed is as follows:

$1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4$

However, only $0.045 mol$ of $K2SO4$ is present. Therefore, $0.045 mol$ of $K2SO4$ reacts with $0.045 mol$ of $BaCl2$ to produce $0.045 mol$ of $BaSO4$ .

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

The number of moles $n$ of a substance is equal to the ratio of the mass $m$ of that substance to the molar mass $MM$ of the substance.

$n=mMM$ â€¦â€¦ (2)

Substitute $n$ as $0.045 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.045 mol=m233.38 g mol−1m=0.045 mol×233.38 g mol−1=10.5021 g$

Thus, the mass of barium sulfate formed is $l0.502l g$ .

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(b)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $100 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $100 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.01 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.1 Ln=0.5 mol L−1×0.1 L=0.05 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The number of moles of $BaSO4$ produced as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(c)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $500 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $500 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.5 Ln=0.5 mol L−1×0.5 L=0.25 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$∴ 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2. ∴ 0.01 moles of BaCl2 reacts with 0.01 moles of K2 SO4 to form 0.01 moles of BaSO4$

$1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2$

However, the number of moles of $BaCl2$ is only $0.01 mol$ .

$0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The amount of $BaSO4$ formed is as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

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Answer 5

Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

(a)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $90 ml$ and of barium chloride solution is $500 ml$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $90 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$90 mL=1 L1000 mL×90 mL=0.09 L$

Convert $500 ml$ volume of barium chloride solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Molarity of the solution $M$ is equal to the ratio of the number of moles $n$ of a solute to the volume $V$ of the solution in litres.

$M=nV$ â€¦â€¦ (1)

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.5 Ln=0.1 mol L−1×0.5 L=0.05 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.09 L$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$0.5 mol L−1=n0.09 Ln=0.5 mol L−1×0.09 L=0.045 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4$

Barium sulfate formed is as follows:

$1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4$

However, only $0.045 mol$ of $K2SO4$ is present. Therefore, $0.045 mol$ of $K2SO4$ reacts with $0.045 mol$ of $BaCl2$ to produce $0.045 mol$ of $BaSO4$ .

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

The number of moles $n$ of a substance is equal to the ratio of the mass $m$ of that substance to the molar mass $MM$ of the substance.

$n=mMM$ â€¦â€¦ (2)

Substitute $n$ as $0.045 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.045 mol=m233.38 g mol−1m=0.045 mol×233.38 g mol−1=10.5021 g$

Thus, the mass of barium sulfate formed is $l0.502l g$ .

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(b)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $100 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $100 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.01 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.1 Ln=0.5 mol L−1×0.1 L=0.05 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The number of moles of $BaSO4$ produced as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(c)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $500 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $500 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.5 Ln=0.5 mol L−1×0.5 L=0.25 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$∴ 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2. ∴ 0.01 moles of BaCl2 reacts with 0.01 moles of K2 SO4 to form 0.01 moles of BaSO4$

$1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2$

However, the number of moles of $BaCl2$ is only $0.01 mol$ .

$0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The amount of $BaSO4$ formed is as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

Answer 6

Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

(a)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $90 ml$ and of barium chloride solution is $500 ml$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $90 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$90 mL=1 L1000 mL×90 mL=0.09 L$

Convert $500 ml$ volume of barium chloride solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Molarity of the solution $M$ is equal to the ratio of the number of moles $n$ of a solute to the volume $V$ of the solution in litres.

$M=nV$ â€¦â€¦ (1)

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.5 Ln=0.1 mol L−1×0.5 L=0.05 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.09 L$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$0.5 mol L−1=n0.09 Ln=0.5 mol L−1×0.09 L=0.045 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4$

Barium sulfate formed is as follows:

$1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4$

However, only $0.045 mol$ of $K2SO4$ is present. Therefore, $0.045 mol$ of $K2SO4$ reacts with $0.045 mol$ of $BaCl2$ to produce $0.045 mol$ of $BaSO4$ .

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

The number of moles $n$ of a substance is equal to the ratio of the mass $m$ of that substance to the molar mass $MM$ of the substance.

$n=mMM$ â€¦â€¦ (2)

Substitute $n$ as $0.045 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.045 mol=m233.38 g mol−1m=0.045 mol×233.38 g mol−1=10.5021 g$

Thus, the mass of barium sulfate formed is $l0.502l g$ .

Answer 7

Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $90 ml$ and of barium chloride solution is $500 ml$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $90 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$90 mL=1 L1000 mL×90 mL=0.09 L$

Convert $500 ml$ volume of barium chloride solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Molarity of the solution $M$ is equal to the ratio of the number of moles $n$ of a solute to the volume $V$ of the solution in litres.

$M=nV$ â€¦â€¦ (1)

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.5 Ln=0.1 mol L−1×0.5 L=0.05 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.09 L$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$0.5 mol L−1=n0.09 Ln=0.5 mol L−1×0.09 L=0.045 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$1 mol BaCl2 reacts with=1 mol K2SO40.05 mol BaCl2 reacts with=1 mol K2SO41 mol BaCl2×0.05 mol BaCl2=0.05 mol K2SO4$

Barium sulfate formed is as follows:

$1 mol K2SO4 forms=1 mol BaSO40.05 mol K2SO4 forms=1 mol BaSO41 mol K2SO4×0.05 mol K2SO4=0.05 mol BaSO4$

However, only $0.045 mol$ of $K2SO4$ is present. Therefore, $0.045 mol$ of $K2SO4$ reacts with $0.045 mol$ of $BaCl2$ to produce $0.045 mol$ of $BaSO4$ .

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

The number of moles $n$ of a substance is equal to the ratio of the mass $m$ of that substance to the molar mass $MM$ of the substance.

$n=mMM$ â€¦â€¦ (2)

Substitute $n$ as $0.045 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.045 mol=m233.38 g mol−1m=0.045 mol×233.38 g mol−1=10.5021 g$

Thus, the mass of barium sulfate formed is $l0.502l g$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(b)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $100 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $100 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.01 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.1 Ln=0.5 mol L−1×0.1 L=0.05 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The number of moles of $BaSO4$ produced as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $100 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $100 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.01 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.1 Ln=0.5 mol L−1×0.1 L=0.05 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$1 mol BaCl2 reacts=1 mol K2SO40.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The number of moles of $BaSO4$ produced as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

(c)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $500 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $500 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.5 Ln=0.5 mol L−1×0.5 L=0.25 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$∴ 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2. ∴ 0.01 moles of BaCl2 reacts with 0.01 moles of K2 SO4 to form 0.01 moles of BaSO4$

$1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2$

However, the number of moles of $BaCl2$ is only $0.01 mol$ .

$0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The amount of $BaSO4$ formed is as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

Answer 19

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Answer 20

(c)

Expert Solution

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100 M$ and of potassium sulfate solution is $0.500 M$ . The volume of potassium sulfate solution is $500 ml$ and of barium chloride solution is $100 ml$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

$BaCl2aq+K2SO4aq→BaSO4s+2KClaq$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

Convert volume units from milliliters to liters as follows:

$1 L=1000 mL1 mL=1 L1000 mL$

Convert $500 ml$ volume of potassium sulfate solution from milliliters to liters as follows:

$500 mL=1 L1000 mL×500 mL=0.5 L$

Convert $100 ml$ volume of barium chloride solution from milliliters to liters as follows:

$100 mL=1 L1000 mL×100 mL=0.1 L$

Substitute $M$ as $0.1 mol L−1$ and $V$ as $0.1 L$ in equation (1) to determine the number of moles of barium chloride as follows:

$0.1 mol L−1=n0.1 Ln=0.1 mol L−1×0.1 L=0.01 moles$

Substitute $M$ as $0.5 mol L−1$ and $V$ as $0.5 L$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$0.5 mol L−1=n0.5 Ln=0.5 mol L−1×0.5 L=0.25 moles$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$∴ 0.25 moles of K2SO4 must react with 0.25 mole of BaCl2 to form 0.25 mole of BaSO4But there are only 0.01 mole of BaCl2. ∴ 0.01 moles of BaCl2 reacts with 0.01 moles of K2 SO4 to form 0.01 moles of BaSO4$

$1 mol BaCl2 reacts=1 mol K2SO40.25 mol K2SO4 reacts=1 mol BaCl21 mol K2SO4×0.25 mol K2SO4=0.25 mol BaCl2$

However, the number of moles of $BaCl2$ is only $0.01 mol$ .

$0.01 mol BaCl2 reacts=1 mol K2SO41 mol BaCl2×0.01 mol BaCl2=0.01 mol K2SO4$

The amount of $BaSO4$ formed is as follows:

$1 mol K2SO4 produces=1 mol BaSO40.01 mol K2SO4 produces=1 mol BaSO41 mol K2SO4×0.01 mol K2SO4=0.01 mol BaSO4$

The molar mass of $BaSO4$ is as follows:

$BaSO4=137.38 g mol−1+32 g mol−1+4×16 g mol−1=233.38 g mol−1$

Substitute $n$ as $0.01 moles$ and $MM$ as $233.38 g mol−1$ in equation (2) to determine the mass of $BaSO4$ as follows:

$0.01 mol=m233.38 g mol−1m=0.01 mol×233.38 g mol−1=2.3338 g$

Thus, the mass of barium sulfate formed is $2.3338 g$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Ch. 11 - Prob. 1QC Ch. 11 - Prob. 2QC Ch. 11 - Prob. 3QC Ch. 11 - Prob. 4QC Ch. 11 - Prob. 5QC Ch. 11 - Prob. 6QC Ch. 11 - Prob. 1PP Ch. 11 - Prob. 2PP Ch. 11 - Prob. 3PP Ch. 11 - Prob. 4PP

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

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The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

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Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

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The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

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Interpretation: The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

Explanation of Solution

Interpretation: The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

Interpretation: The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

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Chapter 11 Solutions

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.