Chapter 9, Problem 57A

Interpretation Introduction

Interpretation: The number of moles and mass in grams produced of each product in the given reaction needs to be calculated when 20.0 g of each pure hydrazine and pure oxygen are used.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

3

Answer to Problem 57A

The number of moles and mass in grams produced of each product in the given reaction is:

  $\begin{array}{l}\text{0}\text{.624 mol and }17.48{\text{ of N}}_{\text{2}}\\ \text{1}\text{.25 mol and }22.52{\text{ g of H}}_{\text{2}}\text{O}\end{array}$

Explanation of Solution

The balanced complete chemical equation for the reaction of hydrazine with oxygen is:

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

The number of moles of both the reactants are calculated using formula:

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

The molar mass of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$ is 32.0452 g/mol and 31.998 g/mol respectively.

Substituting the values:

  $\begin{array}{l}{\text{n}}_{{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{ = }\frac{\text{20}\text{.0 g}}{\text{32}\text{.0452 g/mol}}\\ {\text{n}}_{{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{ = 0}\text{.624 mol}\end{array}$

  $\begin{array}{l}{\text{n}}_{\text{S}}\text{ = }\frac{\text{20}\text{.0 g}}{\text{31}\text{.998 g/mol}}\\ {\text{n}}_{\text{S}}\text{ = 0}\text{.625 mol}\end{array}$

From the balanced chemical reaction, the mole ratio of iron and sulfur is 1:1 so, the limiting reactant (the amount of product formed in a chemical reaction is dependent on the reactant known as limiting reactant, which is totally consumed during completion of reaction) is ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ as it contains less number of moles in comparison to the number of moles of ${\text{O}}_{\text{2}}$ . So, the number of moles of products produced from $\text{0}\text{.624 mol}$ of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ (as ${\text{O}}_{\text{2}}$ is present in excess) is:

  $\text{0}{\text{.624 mol of N}}_{\text{2}}{\text{H}}_{\text{4}}\text{ ×}\frac{{\text{1 mol of N}}_{\text{2}}}{{\text{1 mol of N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{ = 0}{\text{.624 mol of N}}_{\text{2}}$

  $\text{0}{\text{.624 mol of N}}_{\text{2}}{\text{H}}_{\text{4}}\text{ ×}\frac{{\text{2 mol of H}}_{\text{2}}\text{O}}{{\text{1 mol of N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{ = 1}{\text{.25 mol of H}}_{\text{2}}\text{O}$

Now, the mass of each products formed is calculated as:

The molar mass of ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ is 28.01 g/mol and 18.015 g/mol. So,

  $\begin{array}{l}{\text{Number of moles of N}}_{\text{2}}=\frac{{\text{Mass of N}}_{\text{2}}}{{\text{Molar mass of N}}_{\text{2}}}\\ \text{0}\text{.624 mol}=\frac{{\text{Mass of N}}_{\text{2}}}{\text{28}\text{.01 g/mol}}\\ {\text{Mass of N}}_{\text{2}}\text{ = }0.624\text{ mol}\times 28.01\text{ g/mol}\\ {\text{Mass of N}}_{\text{2}}\text{ = }17.48\text{ g}\end{array}$

  $\begin{array}{l}{\text{Number of moles of H}}_{\text{2}}\text{O}=\frac{{\text{Mass of H}}_{\text{2}}\text{O}}{{\text{Molar mass of H}}_{\text{2}}\text{O}}\\ \text{1}\text{.25 mol}=\frac{{\text{Mass of H}}_{\text{2}}\text{O}}{\text{18}\text{.015 g/mol}}\\ {\text{Mass of H}}_{\text{2}}\text{O = }1.25\text{ mol}\times 18.015\text{ g/mol}\\ {\text{Mass of H}}_{\text{2}}\text{O = }22.52\text{ g}\end{array}$

Hence, the number of moles and mass in grams produced of each product in the given reaction is:

  $\begin{array}{l}\text{0}\text{.624 mol and }17.48{\text{ of N}}_{\text{2}}\\ \text{1}\text{.25 mol and }22.52{\text{ g of H}}_{\text{2}}\text{O}\end{array}$