Chapter 12, Problem 54A

Interpretation Introduction

Interpretation: To calculate the percent yield for the given reaction.

Concept Introduction: The ratio of actual yield to theoretical yield is defined as percent yield.

Answer to Problem 54A

The percent yield of this reaction is $\text{91}\text{.1\%}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  ${\text{Sb}}_2{\text{S}}_3\left(s\right)+3\text{Fe}\left(s\right)\to 2\text{ Sb}\left(s\right)+3\text{ FeS}\left(s\right)$

When one mole of an ore of antimony reacts to three moles of iron, it yields two moles of antimony and three moles of iron sulfide.

The molar mass of ${\text{Sb}}_2{\text{S}}_3$ is $339.9{\text{ g mol}}^{-1}$ .

The molar mass of $\text{Sb}$ is $121.8{\text{ g mol}}^{-1}$ .

Calculation for grams of antimony is as follows:

  $15.0\cancel{{\text{g Sb}}_2{\text{S}}_3}\times \frac{1\cancel{{\text{mol Sb}}_2{\text{S}}_3}}{339.9\cancel{{\text{g Sb}}_2{\text{S}}_3}}\times \frac{2\cancel{\text{mol Sb}}}{1\cancel{{\text{mol Sb}}_2{\text{S}}_3}}\times \frac{121.8\text{ g Sb}}{1\cancel{\text{mol Sb}}}=10.8\text{ g Sb}$

The calculation for percent yield is as follows:

  $\begin{array}{c}\text{\%yield}=\frac{\text{9}\text{.84 }\cancel{\text{g Sb}}}{\text{10}\text{.8 }\cancel{\text{g Sb}}}\text{×100\%}\\ =\text{91}\text{.1\%}\end{array}$

Conclusion

Thus, the percent yield of this reaction is $\text{91}\text{.1\%}$ .