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)

The Claisen-Schmidt reaction to prepare dibenzalacetone involves the condensation of acetone with benzaldehyde:

$$ CH_3COCH_3 + 2C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O $$

The molar mass of acetone (CH_3COCH_3) is 12×3 + 6 + 16 = 58 g/mol, that of benzaldehyde (C_6H_5CHO) is 12×7 + 6 + 16 = 106 g/mol, and that of dibenzalacetone (C_{17}H_{14}O) is 12×17 + 14 + 16 = 234 g/mol.

To prepare 351 g of dibenzalacetone:

$$ \text{Moles of dibenzalacetone} = \frac{351}{234} = 1.5 \text{ mol} $$

The available 87 g of acetone corresponds to:

$$ \text{Moles of acetone} = \frac{87}{58} = 1.5 \text{ mol} $$

Since the reaction requires 2 mol of benzaldehyde per mole of acetone:

$$ \text{Moles of benzaldehyde} = 2 \times 1.5 = 3 \text{ mol} $$

Thus, the mass of benzaldehyde needed is:

$$ \text{Mass} = 3 \times 106 = 318 \text{ g} $$

Therefore, 318 g of benzaldehyde is required.