In the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was ______ %.

In the Claisen-Schmidt reaction between benzaldehyde and acetone, represented by $$ 2C_6H_5CHO + CH_3COCH_3 \to C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O $$, if 5.3 g of benzaldehyde are used and 3.51 g of dibenzalacetone are obtained, the percentage yield can be calculated as follows.

The molar mass of benzaldehyde is $$6(12) + 5(1) + 12 + 16 + 1 = 106$$ g/mol, and the molar mass of dibenzalacetone is $$17(12) + 14(1) + 16 = 204 + 14 + 16 = 234$$ g/mol.

The number of moles of benzaldehyde is $$\frac{5.3}{106} = 0.05$$ mol. Since 2 mol of benzaldehyde produce 1 mol of dibenzalacetone, the theoretical yield in moles is $$0.05/2 = 0.025$$ mol.

Multiplying by the molar mass of dibenzalacetone gives the theoretical mass as $$0.025 \times 234 = 5.85$$ g.

Therefore, the percentage yield is $$\frac{3.51}{5.85} \times 100 = 60\%$$, so the yield of the reaction is 60%.