Chapter 9, Problem 61A

Interpretation Introduction

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

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

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

Answer to Problem 61A

The number of moles and mass in grams produced of each product in the 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, ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$ 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.

Putting 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 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, ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ 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:

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