JEE Chemistry 2025 Question with Solution

Given: benzaldehyde used = $5.3 \, \text{g}$, actual mass of dibenzalacetone obtained = $3.51 \, \text{g}$.

Find: percentage yield of the reaction.

The balanced reaction is

$$2C_6H_5CHO + CH_3COCH_3 \rightarrow C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O$$

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

Moles of benzaldehyde used:

$$\text{Moles of benzaldehyde} = \frac{5.3}{106} = 0.05 \, \text{mol}$$

From stoichiometry,

$$0.05 \, \text{mol benzaldehyde} \rightarrow 0.025 \, \text{mol dibenzalacetone}$$

Molar mass of dibenzalacetone = $234 \, \text{g/mol}$.

Therefore, theoretical yield is

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

Now calculate percentage yield:

$$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{3.51}{5.85} \times 100 = 60\%$$

Therefore, the percentage yield of the reaction is $60\%$.

Given: the solution shows two approaches. The first approach states a $1:1$ molar ratio and computes $30\%$, but the second approach uses the balanced reaction stoichiometry and gives the final answer $60\%$.

Find: the correct percentage yield.

The correct balanced reaction requires $2$ moles of benzaldehyde for $1$ mole of dibenzalacetone, so the $1:1$ ratio used in the first approach is incorrect.

Using the correct stoichiometry:

$$\text{Moles of benzaldehyde} = \frac{5.3}{106} = 0.05 \, \text{mol}$$ $$\text{Moles of dibenzalacetone theoretically formed} = \frac{0.05}{2} = 0.025 \, \text{mol}$$ $$\text{Theoretical yield} = 0.025 \times 234 = 5.85 \, \text{g}$$ $$\text{Percentage yield} = \frac{3.51}{5.85} \times 100 = 60\%$$

Therefore, the correct numerical answer is $60$.