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.

  $2NO+O_2\to 2N{O}_2$

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.

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

  $2NO+O_2\to 2N{O}_2$

The conversion factor is −

  $\frac{\text{2 }molesofNO}{\text{1 }moleof{O}_2}or\frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}$

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 −

  $\text{1 }moleofNO\times \frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}=0.5moleof{O}_2$

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.

  $2NO+O_2\to 2N{O}_2$

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.

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 −

  $2NO+O_2\to 2N{O}_2$

The conversion factor is −

  $\frac{\text{2 }molesofNO}{\text{1 }moleof{O}_2}or\frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}$

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 −

  $\text{2 }moleofNO\times \frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}=1moleof{O}_2$

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.

  $2NO+O_2\to 2N{O}_2$

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.

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

  $2NO+O_2\to 2N{O}_2$

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 −

  $numberofmoles=\frac{mass}{molarmass}$

  $numberofmoles(NO)=\frac{10g}{30.01g/mol}=0.333mol$

Number of moles of O₂ can be calculated as

  $numberofmoles(O_2)=\frac{10g}{32.00g/mol}=0.313mol$

The conversion factor is −

  $\frac{\text{2 }molesofNO}{\text{1 }moleof{O}_2}or\frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}$

The amount of oxygen required can be calculated as −

  $\text{0}\text{.33 }moleofNO\times \frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}=0.167moleof{O}_2$

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.

  $2NO+O_2\to 2N{O}_2$

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.

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

  $2NO+O_2\to 2N{O}_2$

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 −

  $numberofmoles=\frac{mass}{molarmass}$

  $numberofmoles(NO)=\frac{28g}{30.01g/mol}=0.933mol$

Number of moles of O₂ can be calculated as −

  $numberofmoles(O_2)=\frac{16g}{32.00g/mol}=0.500mol$

The conversion factor is −

  $\frac{\text{2 }molesofNO}{\text{1 }moleof{O}_2}or\frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}$

The amount of oxygen required can be calculated as −

  $\text{0}\text{.933 }moleofNO\times \frac{\text{1 }moleof{O}_2}{\text{2 }molesofNO}=0.467moleof{O}_2$

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