Chapter 12.3, Problem 37LC

Interpretation Introduction

Interpretation: The mass of ${\text{SO}}_3$ produced when $20.0{\text{ g FeS}}_2$ reacts with $16.0{\text{ g O}}_2$ needs to be calculated.

Concept introduction: The balanced equation is the one that has the same number of atoms of each species in both the reactant and product sides. A mole conversion factor is used to do further stoichiometric calculations using a balanced equation.

Answer to Problem 37LC

The grams of ${\text{SO}}_3$ produced when $20.0{\text{ g FeS}}_2$ reacts with $16.0{\text{ g O}}_2$ is $26.57g$ .

Explanation of Solution

The balanced equation given is:

  $4{\text{FeS}}_{\text{2}}(s)+15{\text{O}}_2(g)\to 2{\text{Fe}}_{\text{2}}{\text{O}}_3(s)+8{\text{SO}}_3(g)$

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

Calculate the moles of ${\text{FeS}}_{\text{2}}$ .

  $\begin{array}{l}\text{Number of moles=}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{ = }\frac{\text{20}\text{.0 g}}{\text{119}\text{.98}\frac{\text{g}}{\text{mol}}}\\ \text{ = 0}\text{.166 mol}\end{array}$

Calculate the moles of ${\text{O}}_{\text{2}}$ .

  $\begin{array}{l}\text{Number of moles=}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{ = }\frac{\text{16}\text{.0 g}}{\text{16}\frac{\text{g}}{\text{mol}}}\\ \text{ = 1 mol}\end{array}$

The mole conversion factor to determine the amount of ${\text{SO}}_{\text{3}}$ .

Here, ${\text{FeS}}_{\text{2}}$ is the limiting reagent so, it will determine the amount of product formed.

  $\frac{{\text{1 mol SO}}_{\text{3}}}{\frac{\text{1}}{\text{2}}{\text{ mol FeS}}_{\text{2}}}\text{×0}{\text{.166 mol FeS}}_{\text{2}}\text{= 0}{\text{.332 mol SO}}_{\text{3}}$

Hence, the grams of ${\text{SO}}_{\text{3}}$ formed is:

  $\text{0}{\text{.332 mol SO}}_{\text{3}}\times 8\text{0}\text{.06}\frac{\text{g}}{\text{mol}}=26.\text{57 g}$

Conclusion

The grams of ${\text{SO}}_3$ produced when $20.0{\text{ g FeS}}_2$ reacts with $16.0{\text{ g O}}_2$ is $26.57g$ .