Chapter 12, Problem 83A

Interpretation Introduction

Interpretation: The grams of sulfur trioxide are to be calculated.

Concept Introduction: The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

When converting between a certain number of moles of one reactant or product and moles of another reactant or product, mole ratios are utilized.

Answer to Problem 83A

The grams of sulfur trioxide are $21.36\text{g}$ .

Explanation of Solution

Consider the chemical reaction:

  $\begin{array}{l}{\text{4FeS}}_{\text{2}}{\text{+11O}}_{\text{2}}\to 2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}{\text{+8SO}}_{\text{2}}\hfill \\ {\text{2SO}}_{\text{2}}{\text{+O}}_{\text{2}} \to 2{\text{SO}}_{\text{3}}\hfill \end{array}$

Mass of ${\text{FeS}}_{\text{2}}$ is $\text{20 g}$ .

Mass of oxygen is $\text{16 g}$ .

Molar mass of oxygen is $16\text{g}$ .

Oxygen is the limiting reagent in the reaction.

The moles of ${\text{SO}}_3$ in the reaction are given as follows:

  $16\cancel{\text{g}{\text{O}}_2}\times \frac{1\text{mol}{\text{O}}_2}{32\cancel{\text{g}{\text{O}}_2}}\times \frac{8\text{mol}{\text{SO}}_3}{15\text{mol}{\text{O}}_2}=0.267\text{mol}{\text{SO}}_3$

The mass of ${\text{SO}}_3$ in the reaction is given as follows:

  $0.267\cancel{{\text{mol SO}}_3}\times \frac{80\text{g}}{1\cancel{\text{mol}{\text{SO}}_3}}=21.36\text{g}$