(60)

Moles of benzaldehyde initially taken $\left(n_{C_6{H}_{5}CHO}\right)$

$=\frac{\text{Mass of }{C}_6{H}_{5}CHO}{\text{Molar mass of }{C}_6{H}_{5}CHO}=\frac{5.3}{106}=0.05 \text{mol}$

For 100% yield of the reaction, 2 mol of benzaldehyde give 1 mol of dibenzalacetone, so 0.05 mol of benzaldehyde give 0.025 mol of dibenzalacetone.

Actual moles of dibenzalacetone obtained

$= \frac{\text{Mass of dibenzalacetone obtained}}{\text{Molar mass of dibenzalacetone}}=\frac{3.51}{234}=0.015 \text{mol}$

Percentage yield

$= \frac{\text{Actual moles of dibenzalacetone obtained}}{\text{Moles of dibenzalacetone for 100\% yield}}\times 100$

$=\frac{0.015}{0.025}\times 100=60\%$

(7)