Concept:Percentage yield requires finding the theoretical maximum yield of product using stoichiometry, and then comparing it against the physically obtained experimental mass.
Formula:$$ \% \text{ Yield} = \left( \frac{\text{Actual}}{\text{Theoretical}} \right) \times 100 $$
Solution:- Equation: \( \text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2 \). Ratio is 1:1.
- Molar mass of \( \text{CuCO}_3 \approx 64 (\text{Cu}) + 12 (\text{C}) + 48 (\text{O}_3) = 124 \text{ g/mol} \).
- Molar mass of CuO \( \approx 64 (\text{Cu}) + 16 (\text{O}) = 80 \text{ g/mol} \).
- Based on the 1:1 ratio, 124g of \( \text{CuCO}_3 \) theoretically yields 80g of CuO.
- We are given 24.8g of reactant. Theoretical yield = \( (80 / 124) \times 24.8 = 16 \text{ g} \).
- Actual yield given is 13.9g.
- \( \% \text{ Yield} = (13.9 / 16.0) \times 100 = 86.875\% \).
Why other options are incorrect:- 81.79%, 83.98%, 89.68%: These specific fractions are generated if one uses slightly different isotopes for Cu (like 63.5) and then miscalculates the final long division, but 86.87% matches the exact algebraic outcome of standard rounding.
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