(a) Interpretation: The mass of carbon dioxide that would produce when 6.00 g of H 3 C 6 H 5 O 7 reacts with 20.0 g of NaHCO 3 is to be calculated. Concept introduction: The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Question

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Answer 1

Question

Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of $H3C6H5O7$ in the equation (1).

$n=6.00 g192.12 g/mol=0.0312 mol$

Therefore, the number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

Substitute the mass and molar mass of $NaHCO3$ in the equation (1).

$n=20.0 g84.01 g/mol=0.2381 mol$

Therefore, the number of moles of $NaHCO3$ present in the reaction mixture is $0.2381 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nH3C6H5O7=nNaHCO33$ …(2)

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nNaHCO3$ in the equation (2).

$nCO2=0.2381 mol3=0.0794 mol$

The required amount of $H3C6H5O7$ is greater than the available amount of $H3C6H5O7$ . Therefore, $H3C6H5O7$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of $H3C6H5O7$ . Therefore, the relation between the number of moles of carbon dioxide and $H3C6H5O7$ is given by the expression shown below.

$nCO2=3nH3C6H5O7$ …(3)

Where,

• $nCO2$ is the number of moles of carbon dioxide.

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the equation (3).

$nCO2=(3)(0.0312 mol)=0.0936 mol$

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$m=(0.0936 mol)(44.01 g/mol)=4.12 g$

Therefore, the mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Conclusion

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00 g$ of $H3C6H5O7$ and $20.0 g$ of $NaHCO3$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nNaHCO3=3nH3C6H5O7$

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the above equation.

$nNaHCO3=(3)(0.0312 mol)=0.0936 mol$

The number of moles of $NaHCO3$ reacted is $0.0936 mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of $NaHCO3$ in the equation (4).

$m=(0.0936 mol)(84.01 g/mol)=7.8633 g$

Therefore, the mass of $NaHCO3$ reacted to produce $4.12 g$ of carbon dioxide is $7.8633 g$ .

The excess mass of $NaHCO3$ is given by the expression as shown below.

$mNaHCO3=m1−m2$

Where,

• $m1$ is the initial mass of $NaHCO3$ present in the reaction mixture.

• $m2$ is the mass of $NaHCO3$ reacted in the reaction.

Substitute the value of $m1$ and $m2$ in the above equation.

$mNaHCO3=20.0 g−7.8633 g=12.1367 g$

Therefore, the mass of $NaHCO3$ present in excess is $12.1367 g$ .

Conclusion

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Interpretation Introduction

(c)

Interpretation:

Whether the name of $Na3C6H5O7$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Expert Solution

Answer to Problem 74E

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Explanation of Solution

The compound $Na3C6H5O7$ is salt.

The cation of $Na3C6H5O7$ is sodium ion $(Na+)$ .

The anion of $Na3C6H5O7$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $(C6H5O73−)$ .

Therefore, the name of $Na3C6H5O7$ is sodium citrate.

Conclusion

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

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Answer 2

Question

Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of $H3C6H5O7$ in the equation (1).

$n=6.00 g192.12 g/mol=0.0312 mol$

Therefore, the number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

Substitute the mass and molar mass of $NaHCO3$ in the equation (1).

$n=20.0 g84.01 g/mol=0.2381 mol$

Therefore, the number of moles of $NaHCO3$ present in the reaction mixture is $0.2381 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nH3C6H5O7=nNaHCO33$ …(2)

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nNaHCO3$ in the equation (2).

$nCO2=0.2381 mol3=0.0794 mol$

The required amount of $H3C6H5O7$ is greater than the available amount of $H3C6H5O7$ . Therefore, $H3C6H5O7$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of $H3C6H5O7$ . Therefore, the relation between the number of moles of carbon dioxide and $H3C6H5O7$ is given by the expression shown below.

$nCO2=3nH3C6H5O7$ …(3)

Where,

• $nCO2$ is the number of moles of carbon dioxide.

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the equation (3).

$nCO2=(3)(0.0312 mol)=0.0936 mol$

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$m=(0.0936 mol)(44.01 g/mol)=4.12 g$

Therefore, the mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Conclusion

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00 g$ of $H3C6H5O7$ and $20.0 g$ of $NaHCO3$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nNaHCO3=3nH3C6H5O7$

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the above equation.

$nNaHCO3=(3)(0.0312 mol)=0.0936 mol$

The number of moles of $NaHCO3$ reacted is $0.0936 mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of $NaHCO3$ in the equation (4).

$m=(0.0936 mol)(84.01 g/mol)=7.8633 g$

Therefore, the mass of $NaHCO3$ reacted to produce $4.12 g$ of carbon dioxide is $7.8633 g$ .

The excess mass of $NaHCO3$ is given by the expression as shown below.

$mNaHCO3=m1−m2$

Where,

• $m1$ is the initial mass of $NaHCO3$ present in the reaction mixture.

• $m2$ is the mass of $NaHCO3$ reacted in the reaction.

Substitute the value of $m1$ and $m2$ in the above equation.

$mNaHCO3=20.0 g−7.8633 g=12.1367 g$

Therefore, the mass of $NaHCO3$ present in excess is $12.1367 g$ .

Conclusion

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Interpretation Introduction

(c)

Interpretation:

Whether the name of $Na3C6H5O7$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Expert Solution

Answer to Problem 74E

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Explanation of Solution

The compound $Na3C6H5O7$ is salt.

The cation of $Na3C6H5O7$ is sodium ion $(Na+)$ .

The anion of $Na3C6H5O7$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $(C6H5O73−)$ .

Therefore, the name of $Na3C6H5O7$ is sodium citrate.

Conclusion

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

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Answer 3

Question

Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of $H3C6H5O7$ in the equation (1).

$n=6.00 g192.12 g/mol=0.0312 mol$

Therefore, the number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

Substitute the mass and molar mass of $NaHCO3$ in the equation (1).

$n=20.0 g84.01 g/mol=0.2381 mol$

Therefore, the number of moles of $NaHCO3$ present in the reaction mixture is $0.2381 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nH3C6H5O7=nNaHCO33$ …(2)

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nNaHCO3$ in the equation (2).

$nCO2=0.2381 mol3=0.0794 mol$

The required amount of $H3C6H5O7$ is greater than the available amount of $H3C6H5O7$ . Therefore, $H3C6H5O7$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of $H3C6H5O7$ . Therefore, the relation between the number of moles of carbon dioxide and $H3C6H5O7$ is given by the expression shown below.

$nCO2=3nH3C6H5O7$ …(3)

Where,

• $nCO2$ is the number of moles of carbon dioxide.

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the equation (3).

$nCO2=(3)(0.0312 mol)=0.0936 mol$

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$m=(0.0936 mol)(44.01 g/mol)=4.12 g$

Therefore, the mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Conclusion

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00 g$ of $H3C6H5O7$ and $20.0 g$ of $NaHCO3$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nNaHCO3=3nH3C6H5O7$

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the above equation.

$nNaHCO3=(3)(0.0312 mol)=0.0936 mol$

The number of moles of $NaHCO3$ reacted is $0.0936 mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of $NaHCO3$ in the equation (4).

$m=(0.0936 mol)(84.01 g/mol)=7.8633 g$

Therefore, the mass of $NaHCO3$ reacted to produce $4.12 g$ of carbon dioxide is $7.8633 g$ .

The excess mass of $NaHCO3$ is given by the expression as shown below.

$mNaHCO3=m1−m2$

Where,

• $m1$ is the initial mass of $NaHCO3$ present in the reaction mixture.

• $m2$ is the mass of $NaHCO3$ reacted in the reaction.

Substitute the value of $m1$ and $m2$ in the above equation.

$mNaHCO3=20.0 g−7.8633 g=12.1367 g$

Therefore, the mass of $NaHCO3$ present in excess is $12.1367 g$ .

Conclusion

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Interpretation Introduction

(c)

Interpretation:

Whether the name of $Na3C6H5O7$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Expert Solution

Answer to Problem 74E

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Explanation of Solution

The compound $Na3C6H5O7$ is salt.

The cation of $Na3C6H5O7$ is sodium ion $(Na+)$ .

The anion of $Na3C6H5O7$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $(C6H5O73−)$ .

Therefore, the name of $Na3C6H5O7$ is sodium citrate.

Conclusion

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

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Answer 4

Question

Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of $H3C6H5O7$ in the equation (1).

$n=6.00 g192.12 g/mol=0.0312 mol$

Therefore, the number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

Substitute the mass and molar mass of $NaHCO3$ in the equation (1).

$n=20.0 g84.01 g/mol=0.2381 mol$

Therefore, the number of moles of $NaHCO3$ present in the reaction mixture is $0.2381 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nH3C6H5O7=nNaHCO33$ …(2)

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nNaHCO3$ in the equation (2).

$nCO2=0.2381 mol3=0.0794 mol$

The required amount of $H3C6H5O7$ is greater than the available amount of $H3C6H5O7$ . Therefore, $H3C6H5O7$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of $H3C6H5O7$ . Therefore, the relation between the number of moles of carbon dioxide and $H3C6H5O7$ is given by the expression shown below.

$nCO2=3nH3C6H5O7$ …(3)

Where,

• $nCO2$ is the number of moles of carbon dioxide.

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the equation (3).

$nCO2=(3)(0.0312 mol)=0.0936 mol$

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$m=(0.0936 mol)(44.01 g/mol)=4.12 g$

Therefore, the mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Conclusion

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00 g$ of $H3C6H5O7$ and $20.0 g$ of $NaHCO3$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nNaHCO3=3nH3C6H5O7$

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the above equation.

$nNaHCO3=(3)(0.0312 mol)=0.0936 mol$

The number of moles of $NaHCO3$ reacted is $0.0936 mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of $NaHCO3$ in the equation (4).

$m=(0.0936 mol)(84.01 g/mol)=7.8633 g$

Therefore, the mass of $NaHCO3$ reacted to produce $4.12 g$ of carbon dioxide is $7.8633 g$ .

The excess mass of $NaHCO3$ is given by the expression as shown below.

$mNaHCO3=m1−m2$

Where,

• $m1$ is the initial mass of $NaHCO3$ present in the reaction mixture.

• $m2$ is the mass of $NaHCO3$ reacted in the reaction.

Substitute the value of $m1$ and $m2$ in the above equation.

$mNaHCO3=20.0 g−7.8633 g=12.1367 g$

Therefore, the mass of $NaHCO3$ present in excess is $12.1367 g$ .

Conclusion

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Interpretation Introduction

(c)

Interpretation:

Whether the name of $Na3C6H5O7$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Expert Solution

Answer to Problem 74E

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Explanation of Solution

The compound $Na3C6H5O7$ is salt.

The cation of $Na3C6H5O7$ is sodium ion $(Na+)$ .

The anion of $Na3C6H5O7$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $(C6H5O73−)$ .

Therefore, the name of $Na3C6H5O7$ is sodium citrate.

Conclusion

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

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Answer 5

Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of $H3C6H5O7$ in the equation (1).

$n=6.00 g192.12 g/mol=0.0312 mol$

Therefore, the number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

Substitute the mass and molar mass of $NaHCO3$ in the equation (1).

$n=20.0 g84.01 g/mol=0.2381 mol$

Therefore, the number of moles of $NaHCO3$ present in the reaction mixture is $0.2381 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nH3C6H5O7=nNaHCO33$ …(2)

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nNaHCO3$ in the equation (2).

$nCO2=0.2381 mol3=0.0794 mol$

The required amount of $H3C6H5O7$ is greater than the available amount of $H3C6H5O7$ . Therefore, $H3C6H5O7$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of $H3C6H5O7$ . Therefore, the relation between the number of moles of carbon dioxide and $H3C6H5O7$ is given by the expression shown below.

$nCO2=3nH3C6H5O7$ …(3)

Where,

• $nCO2$ is the number of moles of carbon dioxide.

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the equation (3).

$nCO2=(3)(0.0312 mol)=0.0936 mol$

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$m=(0.0936 mol)(44.01 g/mol)=4.12 g$

Therefore, the mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Conclusion

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00 g$ of $H3C6H5O7$ and $20.0 g$ of $NaHCO3$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nNaHCO3=3nH3C6H5O7$

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the above equation.

$nNaHCO3=(3)(0.0312 mol)=0.0936 mol$

The number of moles of $NaHCO3$ reacted is $0.0936 mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of $NaHCO3$ in the equation (4).

$m=(0.0936 mol)(84.01 g/mol)=7.8633 g$

Therefore, the mass of $NaHCO3$ reacted to produce $4.12 g$ of carbon dioxide is $7.8633 g$ .

The excess mass of $NaHCO3$ is given by the expression as shown below.

$mNaHCO3=m1−m2$

Where,

• $m1$ is the initial mass of $NaHCO3$ present in the reaction mixture.

• $m2$ is the mass of $NaHCO3$ reacted in the reaction.

Substitute the value of $m1$ and $m2$ in the above equation.

$mNaHCO3=20.0 g−7.8633 g=12.1367 g$

Therefore, the mass of $NaHCO3$ present in excess is $12.1367 g$ .

Conclusion

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Interpretation Introduction

(c)

Interpretation:

Whether the name of $Na3C6H5O7$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Expert Solution

Answer to Problem 74E

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Explanation of Solution

The compound $Na3C6H5O7$ is salt.

The cation of $Na3C6H5O7$ is sodium ion $(Na+)$ .

The anion of $Na3C6H5O7$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $(C6H5O73−)$ .

Therefore, the name of $Na3C6H5O7$ is sodium citrate.

Conclusion

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Answer 6

Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of $H3C6H5O7$ in the equation (1).

$n=6.00 g192.12 g/mol=0.0312 mol$

Therefore, the number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

Substitute the mass and molar mass of $NaHCO3$ in the equation (1).

$n=20.0 g84.01 g/mol=0.2381 mol$

Therefore, the number of moles of $NaHCO3$ present in the reaction mixture is $0.2381 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nH3C6H5O7=nNaHCO33$ …(2)

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nNaHCO3$ in the equation (2).

$nCO2=0.2381 mol3=0.0794 mol$

The required amount of $H3C6H5O7$ is greater than the available amount of $H3C6H5O7$ . Therefore, $H3C6H5O7$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of $H3C6H5O7$ . Therefore, the relation between the number of moles of carbon dioxide and $H3C6H5O7$ is given by the expression shown below.

$nCO2=3nH3C6H5O7$ …(3)

Where,

• $nCO2$ is the number of moles of carbon dioxide.

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the equation (3).

$nCO2=(3)(0.0312 mol)=0.0936 mol$

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$m=(0.0936 mol)(44.01 g/mol)=4.12 g$

Therefore, the mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Conclusion

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Answer 7

Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Answer 8

Expert Solution

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the mass and molar mass of $H3C6H5O7$ in the equation (1).

$n=6.00 g192.12 g/mol=0.0312 mol$

Therefore, the number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

Substitute the mass and molar mass of $NaHCO3$ in the equation (1).

$n=20.0 g84.01 g/mol=0.2381 mol$

Therefore, the number of moles of $NaHCO3$ present in the reaction mixture is $0.2381 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nH3C6H5O7=nNaHCO33$ …(2)

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nNaHCO3$ in the equation (2).

$nCO2=0.2381 mol3=0.0794 mol$

The required amount of $H3C6H5O7$ is greater than the available amount of $H3C6H5O7$ . Therefore, $H3C6H5O7$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of $H3C6H5O7$ . Therefore, the relation between the number of moles of carbon dioxide and $H3C6H5O7$ is given by the expression shown below.

$nCO2=3nH3C6H5O7$ …(3)

Where,

• $nCO2$ is the number of moles of carbon dioxide.

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the equation (3).

$nCO2=(3)(0.0312 mol)=0.0936 mol$

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$m=(0.0936 mol)(44.01 g/mol)=4.12 g$

Therefore, the mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Answer 13

Conclusion

The mass of carbon dioxide that would produce when $6.00 g$ of $H3C6H5O7$ reacts with $20.0 g$ of $NaHCO3$ is $4.12 g$ .

Answer 14

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00 g$ of $H3C6H5O7$ and $20.0 g$ of $NaHCO3$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Expert Solution

Answer to Problem 74E

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nNaHCO3=3nH3C6H5O7$

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the above equation.

$nNaHCO3=(3)(0.0312 mol)=0.0936 mol$

The number of moles of $NaHCO3$ reacted is $0.0936 mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of $NaHCO3$ in the equation (4).

$m=(0.0936 mol)(84.01 g/mol)=7.8633 g$

Therefore, the mass of $NaHCO3$ reacted to produce $4.12 g$ of carbon dioxide is $7.8633 g$ .

The excess mass of $NaHCO3$ is given by the expression as shown below.

$mNaHCO3=m1−m2$

Where,

• $m1$ is the initial mass of $NaHCO3$ present in the reaction mixture.

• $m2$ is the mass of $NaHCO3$ reacted in the reaction.

Substitute the value of $m1$ and $m2$ in the above equation.

$mNaHCO3=20.0 g−7.8633 g=12.1367 g$

Therefore, the mass of $NaHCO3$ present in excess is $12.1367 g$ .

Conclusion

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Answer 15

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00 g$ of $H3C6H5O7$ and $20.0 g$ of $NaHCO3$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Answer 16

Expert Solution

Answer to Problem 74E

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The reaction between $H3C6H5O7$ and $NaHCO3$ is shown below.

$H3C6H5O7(aq)+3NaHCO3(aq)→Na3C6H5O7(aq)+3CO2(g)+3H2O(l)$

The mass of $H3C6H5O7$ reacted is $6.00 g$ .

The mass of $NaHCO3$ reacted is $20.0 g$ .

The molar mass of $H3C6H5O7$ is $192.12 g/mol$ .

The molar mass of $NaHCO3$ is $84.01 g/mol$ .

The molar mass of carbon dioxide is $44.01 g/mol$ .

The number of moles of $H3C6H5O7$ present in the reaction mixture is $0.0312 mol$ .

One mole of $H3C6H5O7$ reacts with three moles of $NaHCO3$ . Therefore, the relation between the number of moles of $H3C6H5O7$ and $NaHCO3$ is given by the expression shown below.

$nNaHCO3=3nH3C6H5O7$

Where,

• $nNaHCO3$ is the number of moles of $NaHCO3$ .

• $nH3C6H5O7$ is the number of moles of $H3C6H5O7$ .

Substitute the value of $nH3C6H5O7$ in the above equation.

$nNaHCO3=(3)(0.0312 mol)=0.0936 mol$

The number of moles of $NaHCO3$ reacted is $0.0936 mol$ .

The number of moles of a substance is given by the expression shown below.

$n=mM$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $m$ .

$m=nM$ …(4)

Substitute the value of molar mass and the number of moles of $NaHCO3$ in the equation (4).

$m=(0.0936 mol)(84.01 g/mol)=7.8633 g$

Therefore, the mass of $NaHCO3$ reacted to produce $4.12 g$ of carbon dioxide is $7.8633 g$ .

The excess mass of $NaHCO3$ is given by the expression as shown below.

$mNaHCO3=m1−m2$

Where,

• $m1$ is the initial mass of $NaHCO3$ present in the reaction mixture.

• $m2$ is the mass of $NaHCO3$ reacted in the reaction.

Substitute the value of $m1$ and $m2$ in the above equation.

$mNaHCO3=20.0 g−7.8633 g=12.1367 g$

Therefore, the mass of $NaHCO3$ present in excess is $12.1367 g$ .

Answer 21

Conclusion

The mass of $NaHCO3$ present in excess is $12.1367 g$ .

Answer 22

Interpretation Introduction

(c)

Interpretation:

Whether the name of $Na3C6H5O7$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Expert Solution

Answer to Problem 74E

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Explanation of Solution

The compound $Na3C6H5O7$ is salt.

The cation of $Na3C6H5O7$ is sodium ion $(Na+)$ .

The anion of $Na3C6H5O7$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $(C6H5O73−)$ .

Therefore, the name of $Na3C6H5O7$ is sodium citrate.

Conclusion

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Answer 23

Interpretation Introduction

(c)

Interpretation:

Whether the name of $Na3C6H5O7$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Answer 24

Expert Solution

Answer to Problem 74E

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The compound $Na3C6H5O7$ is salt.

The cation of $Na3C6H5O7$ is sodium ion $(Na+)$ .

The anion of $Na3C6H5O7$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $(C6H5O73−)$ .

Therefore, the name of $Na3C6H5O7$ is sodium citrate.

Answer 29

Conclusion

The name of $Na3C6H5O7$ can be predicted. The name of $Na3C6H5O7$ is sodium citrate.

Answer 30

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Answer to Problem 74E

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Answer to Problem 74E

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Answer to Problem 74E

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