The limiting reactant and the mass of CO 2 formed in the given reaction are to be determined and calculated. Concept Introduction: The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction . The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

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Answer 1

Question

Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of $CO2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Expert Solution & Answer

Answer to Problem 7PP

Solution:

The limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced when $11 g$ of $Na2CO3$ is added to a solution that contains $11 g$ of $HCl$ .

Explanation of Solution

Given Information: The given reaction:

$Na2CO3s+2HClg→2NaClaq+H2Ol+CO2g$

Moles of $Na2CO3$ is $11 g$ and moles of $HCl$ is $11 g$ .

One mole of $Na2CO3$ consists of $106 g$ of $Na2CO3$ . Therefore,

$106 g Na2CO3=1 mole Na2CO31=1 mole Na2CO3106 g Na2CO3$

For $11 g$ ,

$11 g Na2CO3=1 mole Na2CO3106 g Na2CO3×11 g Na2CO3=0.103 mole Na2CO3$

One mole of $HCl$ consists of $36.5 g$ of $HCl$ . Therefore,

$36.5 g HCl=1 mole HCl1=1 mole HCl36.5 g HCl$

For $11 g$ ,

$11 g HCl=1 mole HCl36.5 g HCl×11 g HCl=0.301 mole HCl$

From the given reaction, two moles of $HCl$ react with one mole of $Na2CO3$ .

The number of moles of $Na2CO3$ required to react with $0.301 moles$ of $HCl$ ,

$0.301 mol of HCl=0.301 mol of HCl×1 mole of Na2CO32 mole of HCl=0.155 mol Na2CO3$

The number of moles required is $0.155 mol$ but the actual number of moles that is provided in the given reaction is $0.103 mol$ .

Thus, $Na2CO3$ is the limiting reactant.

Now, $1 mol$ of $Na2CO3$ gives $1 mol$ or $44 g$ of $CO2$ , then the amount of $CO2$ produced from $0.103 mol$ is :

$1 mol Na2CO3=1 mol CO21=1 mol CO21 mol Na2CO3$

$0.103 mol Na2CO3=1 mol CO21 mol Na2CO3×0.103 mol Na2CO3=0.103 mol CO2$

The expression that shows the relation between the moles and the mass of the substance is:

$n=mM$

Here, $n$ is the number of moles, $M$ is the molar mass, and $m$ is the mass is the substance.

Substitute $0.103 mole$ for $n$ and $44 g/mol$ for $M$ in the above expression.

$0.103 mol=m44 g/molm=4.532 g$

Conclusion

Limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced.

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Answer 2

Question

Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of $CO2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Expert Solution & Answer

Answer to Problem 7PP

Solution:

The limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced when $11 g$ of $Na2CO3$ is added to a solution that contains $11 g$ of $HCl$ .

Explanation of Solution

Given Information: The given reaction:

$Na2CO3s+2HClg→2NaClaq+H2Ol+CO2g$

Moles of $Na2CO3$ is $11 g$ and moles of $HCl$ is $11 g$ .

One mole of $Na2CO3$ consists of $106 g$ of $Na2CO3$ . Therefore,

$106 g Na2CO3=1 mole Na2CO31=1 mole Na2CO3106 g Na2CO3$

For $11 g$ ,

$11 g Na2CO3=1 mole Na2CO3106 g Na2CO3×11 g Na2CO3=0.103 mole Na2CO3$

One mole of $HCl$ consists of $36.5 g$ of $HCl$ . Therefore,

$36.5 g HCl=1 mole HCl1=1 mole HCl36.5 g HCl$

For $11 g$ ,

$11 g HCl=1 mole HCl36.5 g HCl×11 g HCl=0.301 mole HCl$

From the given reaction, two moles of $HCl$ react with one mole of $Na2CO3$ .

The number of moles of $Na2CO3$ required to react with $0.301 moles$ of $HCl$ ,

$0.301 mol of HCl=0.301 mol of HCl×1 mole of Na2CO32 mole of HCl=0.155 mol Na2CO3$

The number of moles required is $0.155 mol$ but the actual number of moles that is provided in the given reaction is $0.103 mol$ .

Thus, $Na2CO3$ is the limiting reactant.

Now, $1 mol$ of $Na2CO3$ gives $1 mol$ or $44 g$ of $CO2$ , then the amount of $CO2$ produced from $0.103 mol$ is :

$1 mol Na2CO3=1 mol CO21=1 mol CO21 mol Na2CO3$

$0.103 mol Na2CO3=1 mol CO21 mol Na2CO3×0.103 mol Na2CO3=0.103 mol CO2$

The expression that shows the relation between the moles and the mass of the substance is:

$n=mM$

Here, $n$ is the number of moles, $M$ is the molar mass, and $m$ is the mass is the substance.

Substitute $0.103 mole$ for $n$ and $44 g/mol$ for $M$ in the above expression.

$0.103 mol=m44 g/molm=4.532 g$

Conclusion

Limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced.

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Answer 3

Question

Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of $CO2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Expert Solution & Answer

Answer to Problem 7PP

Solution:

The limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced when $11 g$ of $Na2CO3$ is added to a solution that contains $11 g$ of $HCl$ .

Explanation of Solution

Given Information: The given reaction:

$Na2CO3s+2HClg→2NaClaq+H2Ol+CO2g$

Moles of $Na2CO3$ is $11 g$ and moles of $HCl$ is $11 g$ .

One mole of $Na2CO3$ consists of $106 g$ of $Na2CO3$ . Therefore,

$106 g Na2CO3=1 mole Na2CO31=1 mole Na2CO3106 g Na2CO3$

For $11 g$ ,

$11 g Na2CO3=1 mole Na2CO3106 g Na2CO3×11 g Na2CO3=0.103 mole Na2CO3$

One mole of $HCl$ consists of $36.5 g$ of $HCl$ . Therefore,

$36.5 g HCl=1 mole HCl1=1 mole HCl36.5 g HCl$

For $11 g$ ,

$11 g HCl=1 mole HCl36.5 g HCl×11 g HCl=0.301 mole HCl$

From the given reaction, two moles of $HCl$ react with one mole of $Na2CO3$ .

The number of moles of $Na2CO3$ required to react with $0.301 moles$ of $HCl$ ,

$0.301 mol of HCl=0.301 mol of HCl×1 mole of Na2CO32 mole of HCl=0.155 mol Na2CO3$

The number of moles required is $0.155 mol$ but the actual number of moles that is provided in the given reaction is $0.103 mol$ .

Thus, $Na2CO3$ is the limiting reactant.

Now, $1 mol$ of $Na2CO3$ gives $1 mol$ or $44 g$ of $CO2$ , then the amount of $CO2$ produced from $0.103 mol$ is :

$1 mol Na2CO3=1 mol CO21=1 mol CO21 mol Na2CO3$

$0.103 mol Na2CO3=1 mol CO21 mol Na2CO3×0.103 mol Na2CO3=0.103 mol CO2$

The expression that shows the relation between the moles and the mass of the substance is:

$n=mM$

Here, $n$ is the number of moles, $M$ is the molar mass, and $m$ is the mass is the substance.

Substitute $0.103 mole$ for $n$ and $44 g/mol$ for $M$ in the above expression.

$0.103 mol=m44 g/molm=4.532 g$

Conclusion

Limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced.

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Answer 4

Question

Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of $CO2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Expert Solution & Answer

Answer to Problem 7PP

Solution:

The limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced when $11 g$ of $Na2CO3$ is added to a solution that contains $11 g$ of $HCl$ .

Explanation of Solution

Given Information: The given reaction:

$Na2CO3s+2HClg→2NaClaq+H2Ol+CO2g$

Moles of $Na2CO3$ is $11 g$ and moles of $HCl$ is $11 g$ .

One mole of $Na2CO3$ consists of $106 g$ of $Na2CO3$ . Therefore,

$106 g Na2CO3=1 mole Na2CO31=1 mole Na2CO3106 g Na2CO3$

For $11 g$ ,

$11 g Na2CO3=1 mole Na2CO3106 g Na2CO3×11 g Na2CO3=0.103 mole Na2CO3$

One mole of $HCl$ consists of $36.5 g$ of $HCl$ . Therefore,

$36.5 g HCl=1 mole HCl1=1 mole HCl36.5 g HCl$

For $11 g$ ,

$11 g HCl=1 mole HCl36.5 g HCl×11 g HCl=0.301 mole HCl$

From the given reaction, two moles of $HCl$ react with one mole of $Na2CO3$ .

The number of moles of $Na2CO3$ required to react with $0.301 moles$ of $HCl$ ,

$0.301 mol of HCl=0.301 mol of HCl×1 mole of Na2CO32 mole of HCl=0.155 mol Na2CO3$

The number of moles required is $0.155 mol$ but the actual number of moles that is provided in the given reaction is $0.103 mol$ .

Thus, $Na2CO3$ is the limiting reactant.

Now, $1 mol$ of $Na2CO3$ gives $1 mol$ or $44 g$ of $CO2$ , then the amount of $CO2$ produced from $0.103 mol$ is :

$1 mol Na2CO3=1 mol CO21=1 mol CO21 mol Na2CO3$

$0.103 mol Na2CO3=1 mol CO21 mol Na2CO3×0.103 mol Na2CO3=0.103 mol CO2$

The expression that shows the relation between the moles and the mass of the substance is:

$n=mM$

Here, $n$ is the number of moles, $M$ is the molar mass, and $m$ is the mass is the substance.

Substitute $0.103 mole$ for $n$ and $44 g/mol$ for $M$ in the above expression.

$0.103 mol=m44 g/molm=4.532 g$

Conclusion

Limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced.

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Answer 5

Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of $CO2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Expert Solution & Answer

Answer to Problem 7PP

Solution:

The limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced when $11 g$ of $Na2CO3$ is added to a solution that contains $11 g$ of $HCl$ .

Explanation of Solution

Given Information: The given reaction:

$Na2CO3s+2HClg→2NaClaq+H2Ol+CO2g$

Moles of $Na2CO3$ is $11 g$ and moles of $HCl$ is $11 g$ .

One mole of $Na2CO3$ consists of $106 g$ of $Na2CO3$ . Therefore,

$106 g Na2CO3=1 mole Na2CO31=1 mole Na2CO3106 g Na2CO3$

For $11 g$ ,

$11 g Na2CO3=1 mole Na2CO3106 g Na2CO3×11 g Na2CO3=0.103 mole Na2CO3$

One mole of $HCl$ consists of $36.5 g$ of $HCl$ . Therefore,

$36.5 g HCl=1 mole HCl1=1 mole HCl36.5 g HCl$

For $11 g$ ,

$11 g HCl=1 mole HCl36.5 g HCl×11 g HCl=0.301 mole HCl$

From the given reaction, two moles of $HCl$ react with one mole of $Na2CO3$ .

The number of moles of $Na2CO3$ required to react with $0.301 moles$ of $HCl$ ,

$0.301 mol of HCl=0.301 mol of HCl×1 mole of Na2CO32 mole of HCl=0.155 mol Na2CO3$

The number of moles required is $0.155 mol$ but the actual number of moles that is provided in the given reaction is $0.103 mol$ .

Thus, $Na2CO3$ is the limiting reactant.

Now, $1 mol$ of $Na2CO3$ gives $1 mol$ or $44 g$ of $CO2$ , then the amount of $CO2$ produced from $0.103 mol$ is :

$1 mol Na2CO3=1 mol CO21=1 mol CO21 mol Na2CO3$

$0.103 mol Na2CO3=1 mol CO21 mol Na2CO3×0.103 mol Na2CO3=0.103 mol CO2$

The expression that shows the relation between the moles and the mass of the substance is:

$n=mM$

Here, $n$ is the number of moles, $M$ is the molar mass, and $m$ is the mass is the substance.

Substitute $0.103 mole$ for $n$ and $44 g/mol$ for $M$ in the above expression.

$0.103 mol=m44 g/molm=4.532 g$

Conclusion

Limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced.

Answer 6

Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of $CO2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Expert Solution & Answer

Answer to Problem 7PP

Solution:

The limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced when $11 g$ of $Na2CO3$ is added to a solution that contains $11 g$ of $HCl$ .

Explanation of Solution

Given Information: The given reaction:

$Na2CO3s+2HClg→2NaClaq+H2Ol+CO2g$

Moles of $Na2CO3$ is $11 g$ and moles of $HCl$ is $11 g$ .

One mole of $Na2CO3$ consists of $106 g$ of $Na2CO3$ . Therefore,

$106 g Na2CO3=1 mole Na2CO31=1 mole Na2CO3106 g Na2CO3$

For $11 g$ ,

$11 g Na2CO3=1 mole Na2CO3106 g Na2CO3×11 g Na2CO3=0.103 mole Na2CO3$

One mole of $HCl$ consists of $36.5 g$ of $HCl$ . Therefore,

$36.5 g HCl=1 mole HCl1=1 mole HCl36.5 g HCl$

For $11 g$ ,

$11 g HCl=1 mole HCl36.5 g HCl×11 g HCl=0.301 mole HCl$

From the given reaction, two moles of $HCl$ react with one mole of $Na2CO3$ .

The number of moles of $Na2CO3$ required to react with $0.301 moles$ of $HCl$ ,

$0.301 mol of HCl=0.301 mol of HCl×1 mole of Na2CO32 mole of HCl=0.155 mol Na2CO3$

The number of moles required is $0.155 mol$ but the actual number of moles that is provided in the given reaction is $0.103 mol$ .

Thus, $Na2CO3$ is the limiting reactant.

Now, $1 mol$ of $Na2CO3$ gives $1 mol$ or $44 g$ of $CO2$ , then the amount of $CO2$ produced from $0.103 mol$ is :

$1 mol Na2CO3=1 mol CO21=1 mol CO21 mol Na2CO3$

$0.103 mol Na2CO3=1 mol CO21 mol Na2CO3×0.103 mol Na2CO3=0.103 mol CO2$

The expression that shows the relation between the moles and the mass of the substance is:

$n=mM$

Here, $n$ is the number of moles, $M$ is the molar mass, and $m$ is the mass is the substance.

Substitute $0.103 mole$ for $n$ and $44 g/mol$ for $M$ in the above expression.

$0.103 mol=m44 g/molm=4.532 g$

Conclusion

Limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced.

Answer 7

Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of $CO2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Answer 8

Expert Solution & Answer

Answer to Problem 7PP

Solution:

The limiting reactant is $Na2CO3$ and $4.532 g$ of $CO2$ gas is produced when $11 g$ of $Na2CO3$ is added to a solution that contains $11 g$ of $HCl$ .