Question

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

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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}

2. Calculate moles of CuSO₄:

3. Balanced equation:

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:

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