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.
X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X =_____ g. (nearest integer) [Given : molar mass (in g mol\(^{-1}\)) C : 12, H : 1, O : 16, N : 14]
Fortification of food with iron is done using $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is _______ (Nearest integer).} (Given : Molar mass of $\mathrm{Fe}, \mathrm{S}$ and O respectively are 56,32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Match List-I with List-II: List-I