To find out how many grams of zinc sulfate (ZnSOâ‚„) will be produced when 0.124 moles of zinc (Zn) reacts with excess lead(IV) sulfate (Pb(SOâ‚„)â‚‚), 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 (ZnSOâ‚„). 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 (ZnSOâ‚„):
The molar mass of zinc sulfate (ZnSOâ‚„) is given as 161.44 g/mol.
4. Calculate the Mass of Zinc Sulfate Produced:
To find the mass of ZnSOâ‚„ 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.
