Chapter 9, Problem 22A

Interpretation Introduction

Interpretation:

The amount of sulfur that would react with 1.25 g of copper needs to be deduced based on the given reaction.

Concept Introduction:

• A chemical reaction is represented in terms of a chemical equation with the reactants on the left and the products on the right.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation, i.e., the stoichiometry gives the number of reactants and products involved in the reaction.
• Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants.

Answer to Problem 22A

Mass of S = 0.630 g

Explanation of Solution

The given reaction is:

  $\text{Cu(s) + S(s)}\to \text{CuS(s)}$

Step 1: Calculate the moles of Cu present:

Mass of Cu present = 1.25 g

Atomic mass of Cu = 63.5 g/mol

The number of moles of Cu will be:

  $\begin{array}{c}\text{Moles of Cu = }\frac{\text{Mass of Cu}}{\text{Atomic Mass Cu}}\end{array}$

        $\begin{array}{c}=\frac{\text{1}\text{.25 g}}{63.5\text{ g/mol}}\end{array}$

        $\begin{array}{c}=0.0197\text{ moles}\end{array}$

Step 2: Calculate the moles of S reacted:

Based on the reaction stoichiometry:

1 mole of Cu will react with 1 mole of S.

Therefore, 0.0197 moles of Cu will react with 0.0197 moles of S.

Step 3: Calculate the mass of S reacted:

Moles of S = 0.0197

Atomic weight of S = 32 g/mol

Mass of S will be:

  $\begin{array}{c}\text{Mass of S = Moles }\times \text{ Molecular weight}\end{array}$

        $\begin{array}{c}=0.0197\text{ moles }\times \text{ 32 g/mol }\end{array}$

        $\begin{array}{c}\text{= 0}\text{.630 g}\end{array}$

Conclusion

Therefore, 0.630 g of S will react with 1.25 g of Cu.