Mass of benzaldehyde = 5.3 g

Mass of product = 3.51 g

Claisen - Schmidt reaction :

Condensation of aromatic aldehydes / ketones (without alpha-hydrogen) with diphatic aldehydes /ketones (with alpha-hydrogen) in the presence of weak base to form $\alpha,\beta$-unsaturated aldehydes/ketones.

Given

benzaldehyde $\to$ dibenzalacetone

Formula of percent yield

Percent yield $$ = {{actual\,yield} \over {Theoretical\,yield}} \times 100$$

Actual yield = 3.51 g

From the reaction, the stoichiometric ratio between benzaldehyde and dibenzal acetone is 2 : 1

benzaldehyde : dibenzalacetone

$$2:1$$

$$1:{1 \over 2}$$

Moles of benzaldehyde:

$$Moles = {{Mass} \over {Molar\,mass}}$$

$$ = {{5.3\,g} \over {106.12\,g/mol}}$$

= 0.04994 mol

Molarmass of benzaldehyde = 106.12 g/mol

For 1 mole of benzaldehyde, $\frac{1}{2}$ moles product (dibenzalacetone) is formed.

So, for 0.04994 mol benzaldehyde, $\frac{1}{2}\times0.04994$ mol dibenzalacetone is formed.

So, moles of dibenzalacetone = 0.02497 mol

Mass of product dibenzalacetone (Theoretical yield)

Mass = moles $\times$ molarmass

$=0.02497$ mol $\times$ $234.29$ g/mol

= 5.850 g

Percent yield $$ = {{3.51\,g} \over {5.85\,g}} \times 100$$

$$ = 0.6 \times 100 = 60\% $$