Chapter 12, Problem 60A

a.

Interpretation Introduction

Interpretation: To calculate the amount of nitrogen in liters.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The formation of products is dependent on the number of reactants. The reactant that is consumed first is known as the limiting reactant, when this reactant is consumed completely, then there will be no more formation of products.

Answer to Problem 60A

The amount of nitrogen in liters is $8.4\times {10}^2\text{ L}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(l\right)+{\text{O}}_2\left(g\right)\to {\text{N}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)$

When hydrazine reacts with oxygen it yields nitrogen gas along with two moles of water.

The amount of hydrazine is $1.0\text{ kg}$ .

The amount of nitrogen using hydrazine can be calculated as follows:

  $1.0\cancel{{\text{kg N}}_{\text{2}}{\text{H}}_{\text{4}}}\times \frac{{10}^3\cancel{{\text{g N}}_{\text{2}}{\text{H}}_{\text{4}}}}{1.0\cancel{{\text{kg N}}_{\text{2}}{\text{H}}_{\text{4}}}}\times \frac{1\cancel{{\text{mol N}}_{\text{2}}{\text{H}}_{\text{4}}}}{32.0\cancel{{\text{g N}}_{\text{2}}{\text{H}}_{\text{4}}}}\times \frac{1\cancel{{\text{mol N}}_{\text{2}}}}{\text{1 }\cancel{{\text{mol N}}_{\text{2}}{\text{H}}_{\text{4}}}}\times \frac{22.4{\text{ L N}}_2}{\text{ 1 }\cancel{{\text{mol N}}_{\text{2}}}}=7.0\times {10}^3{\text{ L N}}_2$

The amount of nitrogen using oxygen can be calculated as follows:

  $1.2\cancel{{\text{kg O}}_{\text{2}}}\times \frac{{10}^3\cancel{{\text{g O}}_{\text{2}}}}{1.0\cancel{{\text{kg O}}_{\text{2}}}}\times \frac{1\cancel{{\text{mol O}}_{\text{2}}}}{32.0\cancel{{\text{g O}}_{\text{2}}}}\times \frac{1\cancel{{\text{mol N}}_{\text{2}}}}{\text{1 }\cancel{{\text{mol O}}_{\text{2}}}}\times \frac{22.4{\text{ L N}}_2}{\text{ 1 }\cancel{{\text{mol N}}_{\text{2}}}}=8.4\times {10}^2{\text{ L N}}_2$

Therefore, the amount of nitrogen is produced less in the case of hydrazine, which means that the reactant that is consumed first is hydrazine and it is a limiting reagent.

b.

Interpretation Introduction

Interpretation: To calculate the amount of excess reagent in grams.

Concept Introduction: A mole is $6.02\times {10}^{23}$ atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction. The coefficients help in determining the conversion factors between the number of moles for two different substances in a chemical equation.

Answer to Problem 60A

The amount of excess remaining reagent i.e., oxygen is $0.21\text{ kg }$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(l\right)+{\text{O}}_2\left(g\right)\to {\text{N}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)$

When hydrazine reacts with oxygen, it yields nitrogen gas along with two moles of water.

Since hydrazine is a limiting reagent thus oxygen behaves as an excess reagent. When the reaction gets stops, oxygen is the reactant that is still present in the reaction.

The amount of oxygen that is present in excess can be calculated as follows:

  $1.0\cancel{{\text{kg N}}_{\text{2}}{\text{H}}_4}\times \frac{{10}^3\cancel{{\text{g N}}_{\text{2}}{\text{H}}_4}}{1.0\cancel{{\text{kg N}}_{\text{2}}{\text{H}}_4}}\times \frac{1\cancel{{\text{mol N}}_{\text{2}}{\text{H}}_4}}{32.0\cancel{{\text{g N}}_{\text{2}}{\text{H}}_4}}\times \frac{1\cancel{{\text{mol O}}_{\text{2}}}}{\text{1 }\cancel{{\text{mol N}}_{\text{2}}{\text{H}}_4}}\times \frac{32.0{\text{ g O}}_2}{\text{ 1 }\cancel{{\text{mol O}}_{\text{2}}}}=1.0\times {10}^3{\text{ g O}}_2\text{ used}$

The amount of oxygen in kg can be shown via the following conversion as follows:

  $1.0\times {10}^3\cancel{{\text{g O}}_{\text{2}}}\times \frac{1.0{\text{ kg O}}_2}{{10}^3\cancel{{\text{g O}}_{\text{2}}}}=1.0{\text{ kg O}}_2$

The excess remaining reagent can be given as follows:

  $1.2{\text{ kg O}}_2-1.0{\text{ kg O}}_2=0.21{\text{ kg O}}_2$