(a) Interpretation: The limiting reactant from 1.0 mol NO and 1.0 mol of O 2 is to be determined by using the following equation. 2 N O + O 2 → 2 N O 2 Concept Introduction: The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Question

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Answer 1

Question

Chapter 5, Problem 5.97P

Interpretation Introduction

(a)

Interpretation:

The limiting reactant from 1.0 mol NO and 1.0 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 1.0 mol

The number of mol of O₂ = 1.0 mol

The amount of oxygen required can be calculated as −

1 mole of NO×1 mole of O22 moles of NO=0.5 mole of O2

The required amount of oxygen is 0.5mol but the given amount of oxygen (1.0 mol) is more than the required (0.5 mol) amount. Therefore, NO is a limiting reactant.

Interpretation Introduction

(b)

Interpretation:

The limiting reactant from 2.0 mol NO and 0.50 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is O₂.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as −

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 2.0 mol

The number of mol of O₂ = 0.50 mol

The amount of oxygen required can be calculated as −

2 mole of NO×1 mole of O22 moles of NO=1 mole of O2

The required amount of oxygen is 1mol but the given amount of oxygen (0.50 mol) is less than the required (1.0 mol) amount. Therefore, O₂ is a limiting reactant.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant for 10.0 g NO and 10.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 10 g

The mass of O₂ = 10 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=10g30.01g/mol=0.333mol

Number of moles of O₂ can be calculated as

number of moles(O2)=10g32.00g/mol=0.313mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.33 mole of NO×1 mole of O22 moles of NO=0.167 mole of O2

The required (0.167 mol) amount of oxygen is less than the given amount (0.313 mol). Therefore, the limiting reactant is NO.

Interpretation Introduction

(d)

Interpretation:

The limiting reactant for 28.0 g NO and 16.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 28.0 g

The mass of O₂ = 16 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=28g30.01g/mol=0.933mol

Number of moles of O₂ can be calculated as −

number of moles(O2)=16g32.00g/mol=0.500mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.933 mole of NO×1 mole of O22 moles of NO=0.467 mole of O2

The required (0.467 mol) amount of oxygen is less than the given amount (0.500 mol). Therefore, the limiting reactant is NO.

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Answer 2

Question

Chapter 5, Problem 5.97P

Interpretation Introduction

(a)

Interpretation:

The limiting reactant from 1.0 mol NO and 1.0 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 1.0 mol

The number of mol of O₂ = 1.0 mol

The amount of oxygen required can be calculated as −

1 mole of NO×1 mole of O22 moles of NO=0.5 mole of O2

The required amount of oxygen is 0.5mol but the given amount of oxygen (1.0 mol) is more than the required (0.5 mol) amount. Therefore, NO is a limiting reactant.

Interpretation Introduction

(b)

Interpretation:

The limiting reactant from 2.0 mol NO and 0.50 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is O₂.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as −

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 2.0 mol

The number of mol of O₂ = 0.50 mol

The amount of oxygen required can be calculated as −

2 mole of NO×1 mole of O22 moles of NO=1 mole of O2

The required amount of oxygen is 1mol but the given amount of oxygen (0.50 mol) is less than the required (1.0 mol) amount. Therefore, O₂ is a limiting reactant.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant for 10.0 g NO and 10.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 10 g

The mass of O₂ = 10 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=10g30.01g/mol=0.333mol

Number of moles of O₂ can be calculated as

number of moles(O2)=10g32.00g/mol=0.313mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.33 mole of NO×1 mole of O22 moles of NO=0.167 mole of O2

The required (0.167 mol) amount of oxygen is less than the given amount (0.313 mol). Therefore, the limiting reactant is NO.

Interpretation Introduction

(d)

Interpretation:

The limiting reactant for 28.0 g NO and 16.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 28.0 g

The mass of O₂ = 16 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=28g30.01g/mol=0.933mol

Number of moles of O₂ can be calculated as −

number of moles(O2)=16g32.00g/mol=0.500mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.933 mole of NO×1 mole of O22 moles of NO=0.467 mole of O2

The required (0.467 mol) amount of oxygen is less than the given amount (0.500 mol). Therefore, the limiting reactant is NO.

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Answer 3

Question

Chapter 5, Problem 5.97P

Interpretation Introduction

(a)

Interpretation:

The limiting reactant from 1.0 mol NO and 1.0 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 1.0 mol

The number of mol of O₂ = 1.0 mol

The amount of oxygen required can be calculated as −

1 mole of NO×1 mole of O22 moles of NO=0.5 mole of O2

The required amount of oxygen is 0.5mol but the given amount of oxygen (1.0 mol) is more than the required (0.5 mol) amount. Therefore, NO is a limiting reactant.

Interpretation Introduction

(b)

Interpretation:

The limiting reactant from 2.0 mol NO and 0.50 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is O₂.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as −

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 2.0 mol

The number of mol of O₂ = 0.50 mol

The amount of oxygen required can be calculated as −

2 mole of NO×1 mole of O22 moles of NO=1 mole of O2

The required amount of oxygen is 1mol but the given amount of oxygen (0.50 mol) is less than the required (1.0 mol) amount. Therefore, O₂ is a limiting reactant.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant for 10.0 g NO and 10.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 10 g

The mass of O₂ = 10 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=10g30.01g/mol=0.333mol

Number of moles of O₂ can be calculated as

number of moles(O2)=10g32.00g/mol=0.313mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.33 mole of NO×1 mole of O22 moles of NO=0.167 mole of O2

The required (0.167 mol) amount of oxygen is less than the given amount (0.313 mol). Therefore, the limiting reactant is NO.

Interpretation Introduction

(d)

Interpretation:

The limiting reactant for 28.0 g NO and 16.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 28.0 g

The mass of O₂ = 16 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=28g30.01g/mol=0.933mol

Number of moles of O₂ can be calculated as −

number of moles(O2)=16g32.00g/mol=0.500mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.933 mole of NO×1 mole of O22 moles of NO=0.467 mole of O2

The required (0.467 mol) amount of oxygen is less than the given amount (0.500 mol). Therefore, the limiting reactant is NO.

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Answer 4

Question

Chapter 5, Problem 5.97P

Interpretation Introduction

(a)

Interpretation:

The limiting reactant from 1.0 mol NO and 1.0 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 1.0 mol

The number of mol of O₂ = 1.0 mol

The amount of oxygen required can be calculated as −

1 mole of NO×1 mole of O22 moles of NO=0.5 mole of O2

The required amount of oxygen is 0.5mol but the given amount of oxygen (1.0 mol) is more than the required (0.5 mol) amount. Therefore, NO is a limiting reactant.

Interpretation Introduction

(b)

Interpretation:

The limiting reactant from 2.0 mol NO and 0.50 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is O₂.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as −

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 2.0 mol

The number of mol of O₂ = 0.50 mol

The amount of oxygen required can be calculated as −

2 mole of NO×1 mole of O22 moles of NO=1 mole of O2

The required amount of oxygen is 1mol but the given amount of oxygen (0.50 mol) is less than the required (1.0 mol) amount. Therefore, O₂ is a limiting reactant.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant for 10.0 g NO and 10.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 10 g

The mass of O₂ = 10 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=10g30.01g/mol=0.333mol

Number of moles of O₂ can be calculated as

number of moles(O2)=10g32.00g/mol=0.313mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.33 mole of NO×1 mole of O22 moles of NO=0.167 mole of O2

The required (0.167 mol) amount of oxygen is less than the given amount (0.313 mol). Therefore, the limiting reactant is NO.

Interpretation Introduction

(d)

Interpretation:

The limiting reactant for 28.0 g NO and 16.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 28.0 g

The mass of O₂ = 16 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=28g30.01g/mol=0.933mol

Number of moles of O₂ can be calculated as −

number of moles(O2)=16g32.00g/mol=0.500mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.933 mole of NO×1 mole of O22 moles of NO=0.467 mole of O2

The required (0.467 mol) amount of oxygen is less than the given amount (0.500 mol). Therefore, the limiting reactant is NO.

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Answer 5

Chapter 5, Problem 5.97P

Interpretation Introduction

(a)

Interpretation:

The limiting reactant from 1.0 mol NO and 1.0 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 1.0 mol

The number of mol of O₂ = 1.0 mol

The amount of oxygen required can be calculated as −

1 mole of NO×1 mole of O22 moles of NO=0.5 mole of O2

The required amount of oxygen is 0.5mol but the given amount of oxygen (1.0 mol) is more than the required (0.5 mol) amount. Therefore, NO is a limiting reactant.

Interpretation Introduction

(b)

Interpretation:

The limiting reactant from 2.0 mol NO and 0.50 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is O₂.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as −

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 2.0 mol

The number of mol of O₂ = 0.50 mol

The amount of oxygen required can be calculated as −

2 mole of NO×1 mole of O22 moles of NO=1 mole of O2

The required amount of oxygen is 1mol but the given amount of oxygen (0.50 mol) is less than the required (1.0 mol) amount. Therefore, O₂ is a limiting reactant.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant for 10.0 g NO and 10.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 10 g

The mass of O₂ = 10 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=10g30.01g/mol=0.333mol

Number of moles of O₂ can be calculated as

number of moles(O2)=10g32.00g/mol=0.313mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.33 mole of NO×1 mole of O22 moles of NO=0.167 mole of O2

The required (0.167 mol) amount of oxygen is less than the given amount (0.313 mol). Therefore, the limiting reactant is NO.

Interpretation Introduction

(d)

Interpretation:

The limiting reactant for 28.0 g NO and 16.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 28.0 g

The mass of O₂ = 16 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=28g30.01g/mol=0.933mol

Number of moles of O₂ can be calculated as −

number of moles(O2)=16g32.00g/mol=0.500mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.933 mole of NO×1 mole of O22 moles of NO=0.467 mole of O2

The required (0.467 mol) amount of oxygen is less than the given amount (0.500 mol). Therefore, the limiting reactant is NO.

Answer 6

Chapter 5, Problem 5.97P

Interpretation Introduction

(a)

Interpretation:

The limiting reactant from 1.0 mol NO and 1.0 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 1.0 mol

The number of mol of O₂ = 1.0 mol

The amount of oxygen required can be calculated as −

1 mole of NO×1 mole of O22 moles of NO=0.5 mole of O2

The required amount of oxygen is 0.5mol but the given amount of oxygen (1.0 mol) is more than the required (0.5 mol) amount. Therefore, NO is a limiting reactant.

Answer 7

Chapter 5, Problem 5.97P

Interpretation Introduction

(a)

Interpretation:

The limiting reactant from 1.0 mol NO and 1.0 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Answer 8

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 1.0 mol

The number of mol of O₂ = 1.0 mol

The amount of oxygen required can be calculated as −

1 mole of NO×1 mole of O22 moles of NO=0.5 mole of O2

The required amount of oxygen is 0.5mol but the given amount of oxygen (1.0 mol) is more than the required (0.5 mol) amount. Therefore, NO is a limiting reactant.

Answer 13

Interpretation Introduction

(b)

Interpretation:

The limiting reactant from 2.0 mol NO and 0.50 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is O₂.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as −

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 2.0 mol

The number of mol of O₂ = 0.50 mol

The amount of oxygen required can be calculated as −

2 mole of NO×1 mole of O22 moles of NO=1 mole of O2

The required amount of oxygen is 1mol but the given amount of oxygen (0.50 mol) is less than the required (1.0 mol) amount. Therefore, O₂ is a limiting reactant.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The limiting reactant from 2.0 mol NO and 0.50 mol of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Answer 15

Expert Solution

Answer to Problem 5.97P

The limiting reactant is O₂.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as −

2NO+O2→2NO2

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

Given that −

The number of mole of NO = 2.0 mol

The number of mol of O₂ = 0.50 mol

The amount of oxygen required can be calculated as −

2 mole of NO×1 mole of O22 moles of NO=1 mole of O2

The required amount of oxygen is 1mol but the given amount of oxygen (0.50 mol) is less than the required (1.0 mol) amount. Therefore, O₂ is a limiting reactant.

Answer 20

Interpretation Introduction

(c)

Interpretation:

The limiting reactant for 10.0 g NO and 10.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 10 g

The mass of O₂ = 10 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=10g30.01g/mol=0.333mol

Number of moles of O₂ can be calculated as

number of moles(O2)=10g32.00g/mol=0.313mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.33 mole of NO×1 mole of O22 moles of NO=0.167 mole of O2

The required (0.167 mol) amount of oxygen is less than the given amount (0.313 mol). Therefore, the limiting reactant is NO.

Answer 21

Interpretation Introduction

(c)

Interpretation:

The limiting reactant for 10.0 g NO and 10.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Answer 22

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 10 g

The mass of O₂ = 10 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=10g30.01g/mol=0.333mol

Number of moles of O₂ can be calculated as

number of moles(O2)=10g32.00g/mol=0.313mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.33 mole of NO×1 mole of O22 moles of NO=0.167 mole of O2

The required (0.167 mol) amount of oxygen is less than the given amount (0.313 mol). Therefore, the limiting reactant is NO.

Answer 27

Interpretation Introduction

(d)

Interpretation:

The limiting reactant for 28.0 g NO and 16.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 28.0 g

The mass of O₂ = 16 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=28g30.01g/mol=0.933mol

Number of moles of O₂ can be calculated as −

number of moles(O2)=16g32.00g/mol=0.500mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.933 mole of NO×1 mole of O22 moles of NO=0.467 mole of O2

The required (0.467 mol) amount of oxygen is less than the given amount (0.500 mol). Therefore, the limiting reactant is NO.

Answer 28

Interpretation Introduction

(d)

Interpretation:

The limiting reactant for 28.0 g NO and 16.0 g of O₂ is to be determined by using the following equation.

2NO+O2→2NO2

Concept Introduction:

The limiting reactant is defined as the reactant which is completely used in the reaction. The amount of the product formed is determined by the limiting reagent.

Answer 29

Expert Solution

Answer to Problem 5.97P

The limiting reactant is NO.

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The reaction between the two moles of NO and one mole of O₂ results in the two moles of nitrous oxide which is represented as

2NO+O2→2NO2

Given that −

The mass of NO = 28.0 g

The mass of O₂ = 16 g

The molar mass of NO = 30.01 g/mol

The molar mass of O₂ = 32.00 g/mol

Number of moles of NO can be calculated as −

number of moles=massmolar mass

number of moles(NO)=28g30.01g/mol=0.933mol

Number of moles of O₂ can be calculated as −

number of moles(O2)=16g32.00g/mol=0.500mol

The conversion factor is −

2 moles of NO1 mole of O2 or 1 mole of O22 moles of NO

The amount of oxygen required can be calculated as −

0.933 mole of NO×1 mole of O22 moles of NO=0.467 mole of O2

The required (0.467 mol) amount of oxygen is less than the given amount (0.500 mol). Therefore, the limiting reactant is NO.

Answer 34

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Answer to Problem 5.97P

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Answer to Problem 5.97P

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Answer to Problem 5.97P

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Answer to Problem 5.97P

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