In the Claisen-Schmidt reaction to prepare 35 | vedclass

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(B) The chemical equation for the Claisen-Schmidt reaction is: $2C_6H_5CHO + CH_3COCH_3 \xrightarrow{NaOH} C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O$ From the

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$318$

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(B) The chemical equation for the Claisen-Schmidt reaction is: $2C_6H_5CHO + CH_3COCH_3 \xrightarrow{NaOH} C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O$ From the stoichiometry,$1 \ mole$ of acetone reacts with $2 \ moles$ of benzaldehyde to produce $1 \ mole$ of dibenzalacetone. Molar mass of dibenzalacetone $(C_{17}H_{14}O) = (17 	imes 12) + (14 	imes 1) + 16 = 204 + 14 + 16 = 234 \ g/mol$. Number of moles of dibenzalacetone produced $= \frac{351 \ g}{234 \ g/mol} = 1.5 \ moles$. According to the stoichiometry,$1 \ mole$ of dibenzalacetone requires $2 \ moles$ of benzaldehyde. Therefore,$1.5 \ moles$ of dibenzalacetone requires $1.5 	imes 2 = 3 \ moles$ of benzaldehyde. Molar mass of benzaldehyde $(C_6H_5CHO) = (7 	imes 12) + (6 	imes 1) + 16 = 84 + 6 + 16 = 106 \ g/mol$. Mass of benzaldehyde required $= 3 \ moles 	imes 106 \ g/mol = 318 \ g$.

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)

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