Chapter 9, Problem 28A

Interpretation Introduction

Interpretation: The mass of ammonia required to react with excess of oxygen to produce same amount of water as prepared from reacting $\text{1}\text{.0 g}$ of methane with excess of oxygen should be determined.

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

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio and used as a conversion factor between reactants and products.

Answer to Problem 28A

The mass of ammonia required to react with excess of oxygen to produce same amount of water as prepared from reacting $\text{1}\text{.0 g}$ of methane with excess of oxygen is $1.41\text{ g}$ .

Explanation of Solution

The complete balanced equation for the reaction between methane and oxygen to produce carbon dioxide and water is:

  ${\text{CH}}_4\left(\text{g}\right){\text{ + 2O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{ + 2H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The number of moles of reactants, ${\text{CH}}_4$ is calculated using formula:

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

The molar mass of ${\text{CH}}_4$ is 16.042 g/mol.

Putting the values:

  $\begin{array}{l}{\text{n}}_{{\text{CH}}_4}\text{ = }\frac{\text{1}\text{.0 g}}{\text{16}\text{.042 g/mol}}\\ {\text{n}}_{{\text{CH}}_4}\text{ = 0}\text{.0623 mol}\end{array}$

From the balanced reaction, the mole ratio between water and methane is 2:1 that is 2 moles of water are produced from 1 mole of methane. So, the number of moles of water produced from $\text{0}\text{.0623 mol}$ of methane is:

  $\text{0}{\text{.0623 mol of CH}}_4\text{ ×}\frac{{\text{2 mol of H}}_{\text{2}}\text{O}}{{\text{1 mol of CH}}_4}\text{ = 0}{\text{.1246 mol of H}}_{\text{2}}\text{O}$

Now, the balanced reaction between ammonia and oxygen to produce nitrogen monoxide and water is:

  ${\text{4NH}}_3{\text{ + 5O}}_{\text{2}}\to {\text{4NO + 6H}}_{\text{2}}\text{O}$

From the balanced reaction, the mole ratio between water and ammonia is 6:4 that is 6 moles of water are produced from 4 moles of ammonia. So, the number of moles of ammonia required to produce $\text{ 0}{\text{.1246 mol of H}}_{\text{2}}\text{O}$ is:

  $\text{0}{\text{.1246 mol of H}}_{\text{2}}\text{O ×}\frac{{\text{4 mol of NH}}_3}{{\text{6 mol of H}}_{\text{2}}\text{O}}\text{ = 0}{\text{.083 mol of NH}}_3$

The mass of ammonia is calculated using formula:

  ${\text{Number of moles of NH}}_3=\frac{{\text{Mass of NH}}_3}{{\text{Molar mass of NH}}_3}$

The molar mass of ${\text{NH}}_3$ is 17.03 g/mol so,

  $\begin{array}{l}\text{0}\text{.083 mol}=\frac{{\text{Mass of NH}}_3}{\text{17}\text{.03 g/mol}}\\ {\text{Mass of NH}}_3\text{ = }0.083\text{ mol}\times 17.03\text{ g/mol}\\ {\text{Mass of NH}}_3\text{ = }1.41\text{ g}\end{array}$

Hence, the mass of ammonia required to react with excess of oxygen to produce same amount of water as prepared from reacting $\text{1}\text{.0 g}$ of methane with excess of oxygen is $1.41\text{ g}$ .