Question

In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is _________g. (Nearest integer)

Answer 1

Correct Answer: 318

The reaction is the Claisen-Schmidt condensation:
\[2\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COCH}_3 \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{CH} = \text{CHCOCH} = \text{CHC}_6\text{H}_5 + \text{H}_2\text{O}.\]
Step 1: Reaction Stoichiometry
The stoichiometry dictates that 2 moles of benzaldehyde (\(\text{C}_6\text{H}_5\text{CHO}\)) react with 1 mole of acetone (\(\text{CH}_3\text{COCH}_3\)) to yield 1 mole of dibenzalacetone (\(\text{C}_6\text{H}_5\text{CH} = \text{CHCOCH} = \text{CHC}_6\text{H}_5\)). With 87 g of acetone used, its molar mass is:
\[\text{Molar mass of } \text{CH}_3\text{COCH}_3 = 12 + (1 \times 3) + 12 + 16 + 1 = 58 \, \text{g/mol}.\]
The moles of acetone are calculated as:
\[\text{Moles of acetone} = \frac{\text{Mass of acetone}}{\text{Molar mass of acetone}} = \frac{87}{58} \approx 1.5 \, \text{moles}.\]
Step 2: Calculate Moles of Benzaldehyde Required
Based on the stoichiometry, 2 moles of benzaldehyde are required per 1 mole of acetone. Therefore, the moles of benzaldehyde needed are:
\[\text{Moles of benzaldehyde} = 2 \times \text{Moles of acetone} = 2 \times 1.5 = 3 \, \text{moles}.\]
Step 3: Calculate Mass of Benzaldehyde Required
The molar mass of benzaldehyde (\(\text{C}_6\text{H}_5\text{CHO}\)) is:
\[\text{Molar mass of benzaldehyde} = (6 \times 12) + (5 \times 1) + 12 + 16 = 106 \, \text{g/mol}.\]
The required mass of benzaldehyde is:
\[\text{Mass of benzaldehyde} = \text{Moles of benzaldehyde} \times \text{Molar mass of benzaldehyde} = 3 \times 106 = 318 \, \text{g}.\]
Step 4: Verification with Product Formation
The reaction yields 1 mole of dibenzalacetone for every 2 moles of benzaldehyde. For 1.5 moles of acetone, 1.5 moles of dibenzalacetone are produced, corresponding to 351 g (as provided).
Final Answer: 318 g.