Concept:Percentage yield measures the efficiency of a reaction by comparing the actual amount of product obtained against the theoretical maximum predicted by stoichiometry.
Formula:$$ \% \text{ Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 $$
Solution:- Reaction: 1-propanol (\( \text{C}_3\text{H}_8\text{O} \)) \( \rightarrow \) 1-bromopropane (\( \text{C}_3\text{H}_7\text{Br} \)).
- Molar mass of 1-propanol = 60 g/mol. Molar mass of 1-bromopropane = 123 g/mol.
- According to stoichiometry, 60g of reactant theoretically produces 123g of product.
- Therefore, 10g of reactant theoretically produces: \( (123 / 60) \times 10 = 20.5 \text{ g} \).
- The Actual yield given is 12g.
- \( \% \text{ Yield} = (12 / 20.5) \times 100 = 58.53\% \).
Why other options are incorrect:- 60%, 90%, 50%: These numbers are incorrect mathematical derivations and do not reflect the true ratio of 12g to 20.5g.
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