Chapter 9, Problem 51A

Interpretation Introduction

Interpretation: The theoretical yield of iron(II) sulfide should be calculated when 5.25 g of iron fillings are combined with 12.7 g of sulfur.

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 51A

The theoretical yield of iron(II) sulfide is $8.26\text{ g}$ .

Explanation of Solution

The balanced complete chemical equation for the formation of iron(II) sulfide from iron and sulfur is:

  $\text{Fe(s) + S(s)}\to \text{FeS(s)}$

The number of moles of both the reactants are calculated using formula:

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

The molar mass of Fe and S is 55.845 g/mol and 32.065 g/mol respectively.

Substituting the values:

  $\begin{array}{l}{\text{n}}_{\text{Fe}}\text{ = }\frac{\text{5}\text{.25 g}}{\text{55}\text{.845 g/mol}}\\ {\text{n}}_{\text{Fe}}\text{ = 0}\text{.094 mol}\end{array}$

  $\begin{array}{l}{\text{n}}_{\text{S}}\text{ = }\frac{\text{12}\text{.7 g}}{\text{32}\text{.065 g/mol}}\\ {\text{n}}_{\text{S}}\text{ = 0}\text{.396 mol}\end{array}$

From the balanced chemical reaction, the mole ratio of iron and sulfur is 1:1 so, the limiting reactant (the amount of product formed in a chemical reaction is dependent on the reactant known as limiting reactant, which is totally consumed during completion of reaction) is iron as it contains less number of moles in comparison to the number of moles of sulfur. So, the number of moles of iron(II) sulfide produced from $\text{0}\text{.094 mol}$ of iron (as sulfur is present in excess) is:

  $\text{0}\text{.094 mol of Fe ×}\frac{\text{1 mol of FeS}}{\text{1 mol of Fe}}\text{ = 0}\text{.094 mol of FeS}$

Now, the theoretical yield of iron(II) sulfide is calculated as:

The molar mass of iron(II) sulfide is 87.92 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{0}\text{.094 mol}=\frac{\text{Mass}}{\text{87}\text{.92 g/mol}}\\ \text{Mass = }0.094\text{ mol}\times 87.92\text{ g/mol}\\ \text{Mass = }8.26\text{ g}\end{array}$

Hence, the theoretical yield of iron(II) sulfide is $8.26\text{ g}$ .