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)
Correct Answer: 318
Approach Solution - 1
The reaction involved 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
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 form 1 mole of dibenzalacetone (\(\text{C}_6\text{H}_5\text{CH} = \text{CHCOCH} = \text{CHC}_6\text{H}_5\)).
Given that 87 g of acetone is used, the molar mass of acetone is:
\[\text{Molar mass of } \text{CH}_3\text{COCH}_3 = 12 + (1 \times 3) + 12 + 16 + 1 = 58 \, \text{g/mol}.\]
\[\text{Moles of acetone} = \frac{\text{Mass of acetone}}{\text{Molar mass of acetone}} = \frac{87}{58} \approx 1.5 \, \text{moles}.\]
Step 2: Calculate moles of benzaldehyde required
From the reaction stoichiometry, 2 moles of benzaldehyde are required for every 1 mole of acetone. Thus, 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 the mass of benzaldehyde required
The molar mass of benzaldehyde (\(\text{C}_6\text{H}_5\text{CHO}\)) is:
\[\text{Molar mass of benzaldehyde} = (6 \times 12) + (5 \times 1) + 12 + 16 = 106 \, \text{g/mol}.\]
\[\text{Mass of benzaldehyde} = \text{Moles of benzaldehyde} \times \text{Molar mass of benzaldehyde} = 3 \times 106 = 318 \, \text{g}.\]
Step 4: Verify with product formation
The reaction produces 1 mole of dibenzalacetone for every 2 moles of benzaldehyde. For 1.5 moles of acetone, 1.5 moles of dibenzalacetone are formed, which corresponds to 351 g (given).
Final Answer: 318 g.
Approach Solution -2
The reaction between acetone and benzaldehyde is as follows:
\[ CH_3COCH_3 + 2C_6H_5CHO \rightarrow C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O \]
Thus, 1 mole of acetone reacts with 2 moles of benzaldehyde to form 1 mole of dibenzalacetone.
Step 2: Calculate moles of acetone given.
Molar mass of acetone (CH₃COCH₃) = 58 g/mol
Given mass = 87 g
\[ \text{Moles of acetone} = \frac{87}{58} = 1.5 \, \text{mol (approximately)} \]
Step 3: Use stoichiometric ratio.
From the balanced equation:
1 mole acetone → 2 moles benzaldehyde
Therefore, moles of benzaldehyde required = 2 × 1.5 = 3.0 mol
Step 4: Calculate mass of benzaldehyde required.
Molar mass of benzaldehyde (C₆H₅CHO) = 106 g/mol
\[ \text{Mass} = 3.0 \times 106 = 318 \, \text{g} \]
Step 5: Final Answer.
The amount of benzaldehyde required is 318 g.
Final Answer: 318 g
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