1. Write the balanced Claisen–Schmidt reaction

Dibenzalacetone is formed by condensation of acetone with benzaldehyde:

$2\,\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COCH}_3 \rightarrow \text{C}_{17}\text{H}_{14}\text{O} + 2\,\text{H}_2\text{O}$

So, $2$ moles of benzaldehyde give $1$ mole of dibenzalacetone.

2. Calculate moles of benzaldehyde used

Molar mass of benzaldehyde, $\text{C}_7\text{H}_6\text{O}$:

$7\times 12 + 6\times 1 + 16 = 84 + 6 + 16 = 106\,\text{g mol}^{-1}$

Given mass of benzaldehyde $= 5.3\,\text{g}$

$n(\text{benzaldehyde}) = \frac{5.3}{106} = 0.05\,\text{mol}$

3. Use stoichiometry to find theoretical moles of dibenzalacetone

From the reaction:

$2\,\text{mol benzaldehyde} \rightarrow 1\,\text{mol dibenzalacetone}$

Therefore,

$n(\text{dibenzalacetone}) = \frac{0.05}{2} = 0.025\,\text{mol}$

4. Calculate molar mass of dibenzalacetone

Formula of dibenzalacetone is $\text{C}_{17}\text{H}_{14}\text{O}$.

$17\times 12 + 14\times 1 + 16 = 204 + 14 + 16 = 234\,\text{g mol}^{-1}$

5. Calculate theoretical mass of dibenzalacetone

$m_{\text{theoretical}} = 0.025 \times 234 = 5.85\,\text{g}$

6. Calculate percentage yield

Actual mass obtained $= 3.51\,\text{g}$

$\%\text{yield} = \frac{3.51}{5.85}\times 100$

$\%\text{yield} = 60\%$

7. Final answer

$\boxed{60}$

This matches the stored correct answer.