If 0.320 moles of zinc reacts with excess lead(IV) sulfate how many grams of zinc sulfate would be produced in the following reaction?

To determine the amount of zinc sulfate produced, we first need to look at the balanced chemical reaction between zinc and lead(IV) sulfate. The reaction is:

[ \text{Zn} + \text{Pb(SO}_4\text{)}_2 \rightarrow \text{ZnSO}_4 + \text{Pb} ]

From the reaction, 1 mole of zinc produces 1 mole of zinc sulfate (ZnSO₄). Therefore, 0.320 moles of zinc will produce 0.320 moles of zinc sulfate. To convert moles to grams, we multiply by the molar mass of zinc sulfate (approximately 161.44 g/mol), resulting in:

[ 0.320 , \text{moles} \times 161.44 , \text{g/mol} \approx 51.84 , \text{grams} ]

So, 51.84 grams of zinc sulfate would be produced.