(a) Interpretation: The stoichiometry concept map is to be drawn and the mass of KHCO 3 decomposed is to be stated. Concept introduction: Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction . It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Question

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Answer 1

Question

Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of $KHCO3$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 1

The mass of $KHCO3$ decomposed is $2.282 g$ .

Explanation of Solution

It is given that $255 mL$ of carbon dioxide is produced at $STP$ .

The given chemical reaction is stated below.

$KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(1)

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 2

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of $KHCO3$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of carbon dioxide gas.

At $STP$ the relation between $KHCO3$ and carbon dioxide gas is shown below.

$1 mole of KHCO3=22.4 L CO2$

Therefore, the number of moles of $KHCO3$ is calculated as shown below.

$Number of moles of KHCO3=255 mL CO2×1 L1000 mL×1 mol CO222.4 L CO2×2 mol KHCO31 mol CO2=0.02276 mol KHCO3≈0.0228 mol KHCO3$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of hydrogen is $1.01 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of KHCO3=(Molar mass of K+Molar mass of H+Molar mass of C+(3×Molar mass of O))=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+(3×16.00 g mol−1)=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+48.00 g mol−1=100.12 g mol−1$ The mass of $KHCO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of KHCO31 mol=100.12 g of KHCO3$

The number of moles of $KHCO3$ is $0.0228 mol$ .

The mass for $0.0228 mol$ of $KHCO3$ is calculated as shown below.

$Mass of KHCO3=0.0228 mol×100.12 g KHCO31 mol KHCO3=2.282 g KHCO3$

Therefore, the mass of $KHCO3$ is $2.282 g$ .

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of $KHCO3$ decomposed is $2.282 g$ .

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the $K2CO3$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 3

The mass of $K2CO3$ produced is $1.57 g$ .

Explanation of Solution

The stoichiometry concept mass is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 4

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of $K2CO3$ .

The molar mass of $KHCO3$ is calculated in part (a) is $100.12 g mol−1$ .

The mass calculated of $KHCO3$ in part (a) is $2.282 g$ .

The number of moles of $K2CO3$ is calculated as shown below.

$Number of moles of K2CO3=2.282 g KHCO3×1 mol KHCO3100.12 g KHCO3×1 mol K2CO32 mol K2CO3=0.01139 mol≈0.0114 mol$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of K2CO3=((2×Molar mass of K)+Molar mass of C+(3×Molar mass of O))=(2×39.10 g mol−1)+12.01 g mol−1+(3×16.00 g mol−1)=78.20 g mol−1+12.01 g mol−1+48.00 g mol−1=138.21 g mol−1$

The mass of $K2CO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of K2CO31 mol=138.21 g of K2CO3$

The number of moles of $K2CO3$ is $0.0114 mol$ .

The mass for $0.0114 mol$ of $K2CO3$ is calculated as shown below.

$Mass of K2CO3=0.0114 mol×138.21 g K2CO31 mol K2CO3=1.57 g K2CO3$

Therefore, the mass of $K2CO3$ is $1.57 g$ .

Conclusion

The mass of $K2CO3$ produced after the reaction is $1.57 g$ .

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Answer 2

Question

Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of $KHCO3$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 1

The mass of $KHCO3$ decomposed is $2.282 g$ .

Explanation of Solution

It is given that $255 mL$ of carbon dioxide is produced at $STP$ .

The given chemical reaction is stated below.

$KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(1)

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 2

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of $KHCO3$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of carbon dioxide gas.

At $STP$ the relation between $KHCO3$ and carbon dioxide gas is shown below.

$1 mole of KHCO3=22.4 L CO2$

Therefore, the number of moles of $KHCO3$ is calculated as shown below.

$Number of moles of KHCO3=255 mL CO2×1 L1000 mL×1 mol CO222.4 L CO2×2 mol KHCO31 mol CO2=0.02276 mol KHCO3≈0.0228 mol KHCO3$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of hydrogen is $1.01 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of KHCO3=(Molar mass of K+Molar mass of H+Molar mass of C+(3×Molar mass of O))=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+(3×16.00 g mol−1)=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+48.00 g mol−1=100.12 g mol−1$ The mass of $KHCO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of KHCO31 mol=100.12 g of KHCO3$

The number of moles of $KHCO3$ is $0.0228 mol$ .

The mass for $0.0228 mol$ of $KHCO3$ is calculated as shown below.

$Mass of KHCO3=0.0228 mol×100.12 g KHCO31 mol KHCO3=2.282 g KHCO3$

Therefore, the mass of $KHCO3$ is $2.282 g$ .

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of $KHCO3$ decomposed is $2.282 g$ .

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the $K2CO3$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 3

The mass of $K2CO3$ produced is $1.57 g$ .

Explanation of Solution

The stoichiometry concept mass is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 4

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of $K2CO3$ .

The molar mass of $KHCO3$ is calculated in part (a) is $100.12 g mol−1$ .

The mass calculated of $KHCO3$ in part (a) is $2.282 g$ .

The number of moles of $K2CO3$ is calculated as shown below.

$Number of moles of K2CO3=2.282 g KHCO3×1 mol KHCO3100.12 g KHCO3×1 mol K2CO32 mol K2CO3=0.01139 mol≈0.0114 mol$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of K2CO3=((2×Molar mass of K)+Molar mass of C+(3×Molar mass of O))=(2×39.10 g mol−1)+12.01 g mol−1+(3×16.00 g mol−1)=78.20 g mol−1+12.01 g mol−1+48.00 g mol−1=138.21 g mol−1$

The mass of $K2CO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of K2CO31 mol=138.21 g of K2CO3$

The number of moles of $K2CO3$ is $0.0114 mol$ .

The mass for $0.0114 mol$ of $K2CO3$ is calculated as shown below.

$Mass of K2CO3=0.0114 mol×138.21 g K2CO31 mol K2CO3=1.57 g K2CO3$

Therefore, the mass of $K2CO3$ is $1.57 g$ .

Conclusion

The mass of $K2CO3$ produced after the reaction is $1.57 g$ .

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Answer 3

Question

Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of $KHCO3$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 1

The mass of $KHCO3$ decomposed is $2.282 g$ .

Explanation of Solution

It is given that $255 mL$ of carbon dioxide is produced at $STP$ .

The given chemical reaction is stated below.

$KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(1)

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 2

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of $KHCO3$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of carbon dioxide gas.

At $STP$ the relation between $KHCO3$ and carbon dioxide gas is shown below.

$1 mole of KHCO3=22.4 L CO2$

Therefore, the number of moles of $KHCO3$ is calculated as shown below.

$Number of moles of KHCO3=255 mL CO2×1 L1000 mL×1 mol CO222.4 L CO2×2 mol KHCO31 mol CO2=0.02276 mol KHCO3≈0.0228 mol KHCO3$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of hydrogen is $1.01 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of KHCO3=(Molar mass of K+Molar mass of H+Molar mass of C+(3×Molar mass of O))=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+(3×16.00 g mol−1)=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+48.00 g mol−1=100.12 g mol−1$ The mass of $KHCO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of KHCO31 mol=100.12 g of KHCO3$

The number of moles of $KHCO3$ is $0.0228 mol$ .

The mass for $0.0228 mol$ of $KHCO3$ is calculated as shown below.

$Mass of KHCO3=0.0228 mol×100.12 g KHCO31 mol KHCO3=2.282 g KHCO3$

Therefore, the mass of $KHCO3$ is $2.282 g$ .

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of $KHCO3$ decomposed is $2.282 g$ .

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the $K2CO3$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 3

The mass of $K2CO3$ produced is $1.57 g$ .

Explanation of Solution

The stoichiometry concept mass is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 4

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of $K2CO3$ .

The molar mass of $KHCO3$ is calculated in part (a) is $100.12 g mol−1$ .

The mass calculated of $KHCO3$ in part (a) is $2.282 g$ .

The number of moles of $K2CO3$ is calculated as shown below.

$Number of moles of K2CO3=2.282 g KHCO3×1 mol KHCO3100.12 g KHCO3×1 mol K2CO32 mol K2CO3=0.01139 mol≈0.0114 mol$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of K2CO3=((2×Molar mass of K)+Molar mass of C+(3×Molar mass of O))=(2×39.10 g mol−1)+12.01 g mol−1+(3×16.00 g mol−1)=78.20 g mol−1+12.01 g mol−1+48.00 g mol−1=138.21 g mol−1$

The mass of $K2CO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of K2CO31 mol=138.21 g of K2CO3$

The number of moles of $K2CO3$ is $0.0114 mol$ .

The mass for $0.0114 mol$ of $K2CO3$ is calculated as shown below.

$Mass of K2CO3=0.0114 mol×138.21 g K2CO31 mol K2CO3=1.57 g K2CO3$

Therefore, the mass of $K2CO3$ is $1.57 g$ .

Conclusion

The mass of $K2CO3$ produced after the reaction is $1.57 g$ .

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Answer 4

Question

Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of $KHCO3$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 1

The mass of $KHCO3$ decomposed is $2.282 g$ .

Explanation of Solution

It is given that $255 mL$ of carbon dioxide is produced at $STP$ .

The given chemical reaction is stated below.

$KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(1)

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 2

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of $KHCO3$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of carbon dioxide gas.

At $STP$ the relation between $KHCO3$ and carbon dioxide gas is shown below.

$1 mole of KHCO3=22.4 L CO2$

Therefore, the number of moles of $KHCO3$ is calculated as shown below.

$Number of moles of KHCO3=255 mL CO2×1 L1000 mL×1 mol CO222.4 L CO2×2 mol KHCO31 mol CO2=0.02276 mol KHCO3≈0.0228 mol KHCO3$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of hydrogen is $1.01 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of KHCO3=(Molar mass of K+Molar mass of H+Molar mass of C+(3×Molar mass of O))=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+(3×16.00 g mol−1)=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+48.00 g mol−1=100.12 g mol−1$ The mass of $KHCO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of KHCO31 mol=100.12 g of KHCO3$

The number of moles of $KHCO3$ is $0.0228 mol$ .

The mass for $0.0228 mol$ of $KHCO3$ is calculated as shown below.

$Mass of KHCO3=0.0228 mol×100.12 g KHCO31 mol KHCO3=2.282 g KHCO3$

Therefore, the mass of $KHCO3$ is $2.282 g$ .

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of $KHCO3$ decomposed is $2.282 g$ .

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the $K2CO3$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 3

The mass of $K2CO3$ produced is $1.57 g$ .

Explanation of Solution

The stoichiometry concept mass is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 4

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of $K2CO3$ .

The molar mass of $KHCO3$ is calculated in part (a) is $100.12 g mol−1$ .

The mass calculated of $KHCO3$ in part (a) is $2.282 g$ .

The number of moles of $K2CO3$ is calculated as shown below.

$Number of moles of K2CO3=2.282 g KHCO3×1 mol KHCO3100.12 g KHCO3×1 mol K2CO32 mol K2CO3=0.01139 mol≈0.0114 mol$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of K2CO3=((2×Molar mass of K)+Molar mass of C+(3×Molar mass of O))=(2×39.10 g mol−1)+12.01 g mol−1+(3×16.00 g mol−1)=78.20 g mol−1+12.01 g mol−1+48.00 g mol−1=138.21 g mol−1$

The mass of $K2CO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of K2CO31 mol=138.21 g of K2CO3$

The number of moles of $K2CO3$ is $0.0114 mol$ .

The mass for $0.0114 mol$ of $K2CO3$ is calculated as shown below.

$Mass of K2CO3=0.0114 mol×138.21 g K2CO31 mol K2CO3=1.57 g K2CO3$

Therefore, the mass of $K2CO3$ is $1.57 g$ .

Conclusion

The mass of $K2CO3$ produced after the reaction is $1.57 g$ .

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Answer 5

Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of $KHCO3$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 1

The mass of $KHCO3$ decomposed is $2.282 g$ .

Explanation of Solution

It is given that $255 mL$ of carbon dioxide is produced at $STP$ .

The given chemical reaction is stated below.

$KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(1)

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 2

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of $KHCO3$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of carbon dioxide gas.

At $STP$ the relation between $KHCO3$ and carbon dioxide gas is shown below.

$1 mole of KHCO3=22.4 L CO2$

Therefore, the number of moles of $KHCO3$ is calculated as shown below.

$Number of moles of KHCO3=255 mL CO2×1 L1000 mL×1 mol CO222.4 L CO2×2 mol KHCO31 mol CO2=0.02276 mol KHCO3≈0.0228 mol KHCO3$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of hydrogen is $1.01 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of KHCO3=(Molar mass of K+Molar mass of H+Molar mass of C+(3×Molar mass of O))=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+(3×16.00 g mol−1)=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+48.00 g mol−1=100.12 g mol−1$ The mass of $KHCO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of KHCO31 mol=100.12 g of KHCO3$

The number of moles of $KHCO3$ is $0.0228 mol$ .

The mass for $0.0228 mol$ of $KHCO3$ is calculated as shown below.

$Mass of KHCO3=0.0228 mol×100.12 g KHCO31 mol KHCO3=2.282 g KHCO3$

Therefore, the mass of $KHCO3$ is $2.282 g$ .

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of $KHCO3$ decomposed is $2.282 g$ .

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the $K2CO3$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 3

The mass of $K2CO3$ produced is $1.57 g$ .

Explanation of Solution

The stoichiometry concept mass is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 4

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of $K2CO3$ .

The molar mass of $KHCO3$ is calculated in part (a) is $100.12 g mol−1$ .

The mass calculated of $KHCO3$ in part (a) is $2.282 g$ .

The number of moles of $K2CO3$ is calculated as shown below.

$Number of moles of K2CO3=2.282 g KHCO3×1 mol KHCO3100.12 g KHCO3×1 mol K2CO32 mol K2CO3=0.01139 mol≈0.0114 mol$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of K2CO3=((2×Molar mass of K)+Molar mass of C+(3×Molar mass of O))=(2×39.10 g mol−1)+12.01 g mol−1+(3×16.00 g mol−1)=78.20 g mol−1+12.01 g mol−1+48.00 g mol−1=138.21 g mol−1$

The mass of $K2CO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of K2CO31 mol=138.21 g of K2CO3$

The number of moles of $K2CO3$ is $0.0114 mol$ .

The mass for $0.0114 mol$ of $K2CO3$ is calculated as shown below.

$Mass of K2CO3=0.0114 mol×138.21 g K2CO31 mol K2CO3=1.57 g K2CO3$

Therefore, the mass of $K2CO3$ is $1.57 g$ .

Conclusion

The mass of $K2CO3$ produced after the reaction is $1.57 g$ .

Answer 6

Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of $KHCO3$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 1

The mass of $KHCO3$ decomposed is $2.282 g$ .

Explanation of Solution

It is given that $255 mL$ of carbon dioxide is produced at $STP$ .

The given chemical reaction is stated below.

$KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(1)

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 2

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of $KHCO3$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of carbon dioxide gas.

At $STP$ the relation between $KHCO3$ and carbon dioxide gas is shown below.

$1 mole of KHCO3=22.4 L CO2$

Therefore, the number of moles of $KHCO3$ is calculated as shown below.

$Number of moles of KHCO3=255 mL CO2×1 L1000 mL×1 mol CO222.4 L CO2×2 mol KHCO31 mol CO2=0.02276 mol KHCO3≈0.0228 mol KHCO3$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of hydrogen is $1.01 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of KHCO3=(Molar mass of K+Molar mass of H+Molar mass of C+(3×Molar mass of O))=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+(3×16.00 g mol−1)=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+48.00 g mol−1=100.12 g mol−1$ The mass of $KHCO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of KHCO31 mol=100.12 g of KHCO3$

The number of moles of $KHCO3$ is $0.0228 mol$ .

The mass for $0.0228 mol$ of $KHCO3$ is calculated as shown below.

$Mass of KHCO3=0.0228 mol×100.12 g KHCO31 mol KHCO3=2.282 g KHCO3$

Therefore, the mass of $KHCO3$ is $2.282 g$ .

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of $KHCO3$ decomposed is $2.282 g$ .

Answer 7

Chapter 15, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of $KHCO3$ decomposed is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Answer 8

Expert Solution

Answer to Problem 26E

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 1

The mass of $KHCO3$ decomposed is $2.282 g$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given that $255 mL$ of carbon dioxide is produced at $STP$ .

The given chemical reaction is stated below.

$KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(1)

The stoichiometry concept map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 2

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of $KHCO3$ decomposed.

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of carbon dioxide gas.

At $STP$ the relation between $KHCO3$ and carbon dioxide gas is shown below.

$1 mole of KHCO3=22.4 L CO2$

Therefore, the number of moles of $KHCO3$ is calculated as shown below.

$Number of moles of KHCO3=255 mL CO2×1 L1000 mL×1 mol CO222.4 L CO2×2 mol KHCO31 mol CO2=0.02276 mol KHCO3≈0.0228 mol KHCO3$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of hydrogen is $1.01 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of KHCO3=(Molar mass of K+Molar mass of H+Molar mass of C+(3×Molar mass of O))=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+(3×16.00 g mol−1)=39.10 g mol−1+1.01 g mol−1+12.01 g mol−1+48.00 g mol−1=100.12 g mol−1$ The mass of $KHCO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of KHCO31 mol=100.12 g of KHCO3$

The number of moles of $KHCO3$ is $0.0228 mol$ .

The mass for $0.0228 mol$ of $KHCO3$ is calculated as shown below.

$Mass of KHCO3=0.0228 mol×100.12 g KHCO31 mol KHCO3=2.282 g KHCO3$

Therefore, the mass of $KHCO3$ is $2.282 g$ .

Answer 13

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of $KHCO3$ decomposed is $2.282 g$ .

Answer 14

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the $K2CO3$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Expert Solution

Answer to Problem 26E

The stoichiometry map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 3

The mass of $K2CO3$ produced is $1.57 g$ .

Explanation of Solution

The stoichiometry concept mass is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 4

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of $K2CO3$ .

The molar mass of $KHCO3$ is calculated in part (a) is $100.12 g mol−1$ .

The mass calculated of $KHCO3$ in part (a) is $2.282 g$ .

The number of moles of $K2CO3$ is calculated as shown below.

$Number of moles of K2CO3=2.282 g KHCO3×1 mol KHCO3100.12 g KHCO3×1 mol K2CO32 mol K2CO3=0.01139 mol≈0.0114 mol$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of K2CO3=((2×Molar mass of K)+Molar mass of C+(3×Molar mass of O))=(2×39.10 g mol−1)+12.01 g mol−1+(3×16.00 g mol−1)=78.20 g mol−1+12.01 g mol−1+48.00 g mol−1=138.21 g mol−1$

The mass of $K2CO3$ gas is calculated from the relation as shown below.

$1 mol=Molar mass of K2CO31 mol=138.21 g of K2CO3$

The number of moles of $K2CO3$ is $0.0114 mol$ .

The mass for $0.0114 mol$ of $K2CO3$ is calculated as shown below.

$Mass of K2CO3=0.0114 mol×138.21 g K2CO31 mol K2CO3=1.57 g K2CO3$

Therefore, the mass of $K2CO3$ is $1.57 g$ .

Conclusion

The mass of $K2CO3$ produced after the reaction is $1.57 g$ .

Answer 15

Interpretation Introduction

(b)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of the $K2CO3$ produced is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Answer 16

Expert Solution

Answer to Problem 26E

The stoichiometry map is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 3

The mass of $K2CO3$ produced is $1.57 g$ .

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The stoichiometry concept mass is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 15, Problem 26E , additional homework tip 4

Figure 1

The number of potassium atom and the hydrogen atom are not equal on both sides. A coefficient of $2$ is added in front of $KHCO3$ to balance the chemical reaction.

The balanced equation is stated below.

$2KHCO3(s)→K2CO3(aq)+H2O(l)+CO2(g)$ …(2)

The equation (2) shows that $2$ mole of $KHCO3$ is consumed to form $1$ mole of $K2CO3$ .

The molar mass of $KHCO3$ is calculated in part (a) is $100.12 g mol−1$ .

The mass calculated of $KHCO3$ in part (a) is $2.282 g$ .

The number of moles of $K2CO3$ is calculated as shown below.

$Number of moles of K2CO3=2.282 g KHCO3×1 mol KHCO3100.12 g KHCO3×1 mol K2CO32 mol K2CO3=0.01139 mol≈0.0114 mol$

The molar mass of potassium is $39.10 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of carbon is $12.01 g mol−1$ .

$Total molar mass of K2CO3=((2×Molar mass of K)+Molar mass of C+(3×Molar mass of O))=(2×39.10 g mol−1)+12.01 g mol−1+(3×16.00 g mol−1)=78.20 g mol−1+12.01 g mol−1+48.00 g mol−1=138.21 g mol−1$