Concept:The oxidation of a secondary alcohol (2-propanol) yields a ketone (propanone/acetone). We must calculate theoretical mass based on stoichiometry and then apply the percentage yield to find the actual mass collected.
Formula:$$ \text{Actual Yield} = \text{Theoretical Yield} \times \left( \frac{\% \text{ yield}}{100} \right) $$
Solution:- Molar mass of 2-propanol (\( \text{C}_3\text{H}_8\text{O} \)) = \( (3 \times 12) + 8 + 16 = 60 \text{ g/mol} \).
- Molar mass of propanone (\( \text{C}_3\text{H}_6\text{O} \)) = \( (3 \times 12) + 6 + 16 = 58 \text{ g/mol} \).
- Reaction ratio is 1:1. 60g of reactant theoretically produces 58g of product.
- Given reactant mass is 30g (which is exactly half of 60g).
- Theoretical yield = \( 58 / 2 = 29 \text{ g} \).
- The actual process was only 75% efficient.
- Actual mass produced = \( 29 \times 0.75 = 21.75 \text{ g} \).
Why other options are incorrect:- 29g: This is the absolute maximum theoretical yield if efficiency was 100%.
- 1.74g & 2.74g: Result from severe decimal shifting or incorrect mole substitutions.
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