assuming you used exactly 2.000 g of Zn and 1.26 g of CuSO4 compute the theoretical yeild in grams of copper.

1. Calculate moles of Zn: \frac{2.000 \text{ g}}{65.38 \text{ g/mol}} = 0.0306 \text{ mol} \text{ of Zn}
2. Calculate moles of CuSO₄: \frac{1.26 \text{ g}}{159.61 \text{ g/mol}} = 0.0079 \text{ mol} \text{ of CuSO₄}
3. Balanced equation: \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}
4. Limiting reactant is CuSO₄ because it has fewer moles.
5. Moles of copper produced: 0.0079 mol (same as moles of CuSO₄).
6. Mass of copper produced: 0.0079 \text{ mol} \times 63.55 \text{ g/mol} = 0.50 \text{ g of Cu}