Chapter 9, Problem 5STP

Interpretation Introduction

Interpretation: The mass of waterproduced from $5.0\text{ g}$ of methane 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 5STP

The correct option is (c): $11.2\text{ g}$ .

Explanation of Solution

The given balanced reaction is:

  ${\text{CH}}_4{\text{(g) + 2O}}_{\text{2}}\text{(g)}\to {\text{CO}}_2{\text{(g) + 2H}}_2\text{O}(\text{g)}$

The number of moles of ${\text{CH}}_4$ is calculated as shown:

The molar mass of ${\text{CH}}_4$ is 16.04g/mol.

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{5.0\text{ g}}{\text{16}\text{.04 g/mol}}\\ \text{Number of moles}=\text{ 0}\text{.312 mol}\end{array}$

From the balanced chemical reaction, the mole ratio of ${\text{CH}}_4$ and ${\text{H}}_2\text{O}$ is 1:2 so, the number of moles of ${\text{H}}_2\text{O}$ produced from $\text{0}\text{.312 mol}$ of ${\text{CH}}_4$ is:

  $\text{0}{\text{.312 mol of CH}}_4\text{×}\frac{{\text{2 mol of H}}_2\text{O}}{{\text{1 mol of CH}}_4}\text{ = 0}{\text{.624 mol of H}}_2\text{O}$

Now, the mass of ${\text{H}}_2\text{O}$ produced from $5.0\text{ g}$ of methane is calculated as:

The molar mass of ${\text{H}}_2\text{O}$ is 18.015 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{0}\text{.624 mol }=\frac{\text{Mass}}{18.015\text{ g/mol}}\\ \text{Mass = 0}\text{.624 mol }\times \text{ 18}\text{.015 g/mol}\\ \text{Mass = }11.2\text{ g}\end{array}$

Hence, the mass of water produced from $5.0\text{ g}$ of methaneis $11.2\text{ g}$ .