Chapter 9, Problem 18A

Interpretation Introduction

Interpretation: The mass of water vapor produced from burning of $56.0\text{ g}$ hydrogen gas in air needs to be calculated.

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 18A

The mass of water vapor produced from burning of $56.0\text{ g}$ hydrogen gas in airis $500.46\text{ g}$ .

Explanation of Solution

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

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

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

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{56.0\text{ g}}{\text{2}\text{.016 g/mol}}\\ \text{Number of moles}=27.78\text{ mol}\end{array}$

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 water produced from $27.78\text{ mol}$ of hydrogen is:

  $27.78{\text{ mol of H}}_{\text{2}}\text{×}\frac{{\text{1 mol of H}}_{\text{2}}\text{O}}{{\text{1 mol of H}}_{\text{2}}}\text{ = }27.78{\text{ mol of H}}_{\text{2}}\text{O}$

Now, the mass of water vapor produced from $56.0\text{ g}$ of hydrogen gas is calculated as:

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

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 27.78\text{ mol}=\frac{\text{Mass}}{18.015\text{ g/mol}}\\ \text{Mass = }27.78\text{ mol}\times 18.015\text{ g/mol}\\ \text{Mass = }500.46\text{ g}\end{array}$

Hence, the mass of water vapor produced from burning of $56.0\text{ g}$ hydrogen gas in airis $500.46\text{ g}$ .