Consider the reaction:
[tex]\[
Pb \left( SO _4\right)_2 + 2 Zn \rightarrow 2 ZnSO _4 + Pb
\][/tex]

If [tex]$0.124 \, \text{mol}$[/tex] of zinc reacts with excess lead(IV) sulfate, how many grams of zinc sulfate will be produced in the reaction?

Mass: [tex]$\square \, \text{g} \, \text{ZnSO}_4$[/tex]



Answer :

To find out how many grams of zinc sulfate (ZnSO₄) will be produced when 0.124 moles of zinc (Zn) reacts with excess lead(IV) sulfate (Pb(SO₄)₂), we can follow these detailed steps:

1. Identify the Reaction and Molar Ratios:

From the balanced chemical equation:
[tex]\[ Pb(SO_4)_2 + 2 Zn \rightarrow 2 ZnSO_4 + Pb \][/tex]

We see that 2 moles of zinc (Zn) produce 2 moles of zinc sulfate (ZnSO₄). So, the molar ratio between zinc and zinc sulfate is 1:1.

2. Determine the Moles of Zinc Sulfate Produced:

Given that 0.124 moles of zinc react, and based on the 1:1 molar ratio, this means 0.124 moles of zinc will produce 0.124 moles of zinc sulfate.

3. Calculate the Molar Mass of Zinc Sulfate (ZnSO₄):

The molar mass of zinc sulfate (ZnSO₄) is given as 161.44 g/mol.

4. Calculate the Mass of Zinc Sulfate Produced:

To find the mass of ZnSO₄ produced, use the formula:
[tex]\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \][/tex]

Substituting the values:
[tex]\[ \text{Mass of ZnSO}_4 = 0.124 \, \text{mol} \times 161.44 \, \text{g/mol} \][/tex]

When you multiply these numbers, you get:
[tex]\[ \text{Mass of ZnSO}_4 = 20.01856 \, \text{g} \][/tex]

Therefore, 20.01856 grams of zinc sulfate will be produced in the reaction.