Chapter 12, Problem 58A

Interpretation Introduction

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

Concept Introduction: Percent yield can be calculated by dividing the actual yield to the theoretical yield multiplied by 100.

Answer to Problem 58A

The percent yield of this reaction is $87.1\text{\%}$ .

Explanation of Solution

The given chemical reaction is depicted as follows:

  $4{\text{ FeCO}}_3\left(s\right)+{\text{O}}_2\left(g\right)\to 2{\text{ Fe}}_2{\text{O}}_3\left(s\right)+4{\text{CO}}_2\left(g\right)$

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 ${\text{FeCO}}_3$ is $\text{115}{\text{.85 g mol}}^{-1}$ .

The molar mass of ${\text{Fe}}_2{\text{O}}_3$ is $159.69{\text{ g mol}}^{-1}$ .

Calculation for grams of antimony is as follows:

  $\text{75}\text{.0 }\cancel{{\text{g FeCO}}_3}\times \frac{1\cancel{{\text{mol FeCO}}_3}}{115.85\cancel{{\text{g FeCO}}_3}}\times \frac{2\cancel{{\text{mol Fe}}_2{\text{O}}_3}}{4\cancel{{\text{mol FeCO}}_3}}\times \frac{159.69{\text{ g Fe}}_2{\text{O}}_3}{1\cancel{{\text{mol Fe}}_2{\text{O}}_3}}=51.66{\text{ g Fe}}_2{\text{O}}_3$

The calculation for percent yield is as follows:

  $\begin{array}{c}\text{\%yield}=\frac{\text{45}\text{.0 }\cancel{{\text{g Fe}}_2{\text{O}}_3}}{\text{51}\text{.66 }\cancel{{\text{g Fe}}_2{\text{O}}_3}}\text{×100\%}\\ =87.1\text{\%}\end{array}$

Conclusion

Thus, the percent yield of this reaction is $87.1\text{\%}$ .