Chapter 9.2, Problem 5RQ

Interpretation Introduction

Interpretation: The mass of waterproduced from $3.5\times {10}^2\text{ g}$ of hydrogen gas needs to be calculated when oxygen is present in excess.

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 5RQ

The mass of water produced from $3.5\times {10}^2\text{ g}$ of hydrogen gas is $3.13\times {10}^3\text{ g}$ .

Explanation of Solution

The balanced complete chemical equation for the formation of water from hydrogen and oxygen gas is:

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

The molar mass of hydrogen gas is 2.016 g/mol. The number of moles of hydrogen is calculated as shown:

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{3.5\times {10}^2\text{ g}}{\text{2}\text{.016 g/mol}}\\ \text{Number of moles}=1.74\times {10}^2\text{ mol}\end{array}$

From the balanced chemical reaction, the mole ratio of hydrogen and water is 2:2 that is 1:1 so, the number of moles of waterproduced from $1.74\times {10}^2\text{ mol}$ of hydrogen (as oxygen is present in excess) is:

  $\text{1}{\text{.74×10}}^{\text{2}}{\text{ mol of H}}_{\text{2}}\text{×}\frac{{\text{1 mol of H}}_{\text{2}}\text{O}}{{\text{1 mol of H}}_{\text{2}}}\text{ = 1}{\text{.74×10}}^{\text{2}}{\text{ mol of H}}_{\text{2}}\text{O}$

Now, the mass of water produced from $3.5\times {10}^2\text{ g}$ of hydrogen gasis calculated as:

The molar mass of water is 18.015 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 1.74\times {10}^2\text{ mol}=\frac{\text{Mass}}{18.015\text{ g/mol}}\\ \text{Mass = }1.74\times {10}^2\text{ mol}\times 18.015\text{ g/mol}\\ \text{Mass = }3.13\times {10}^3\text{ g}\end{array}$

Hence, the mass of water produced from $3.5\times {10}^2\text{ g}$ of hydrogen gas is $3.13\times {10}^3\text{ g}$ .