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Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 29 January, Evening Shift — Question 14

In the Claisen–Schmidt reaction to prepare dibenzalacetone from 5.3 g5.3\,\mathrm{g} benzaldehyde, a total of 3.51 g3.51\,\mathrm{g} of product was obtained. The percentage yield in this reaction was

Answer: 60

Numerical answer — enter this value.

Step-by-step solution

Reaction stoichiometry: 2 C6H5CHO+CH3COCH3→C17H14O+2 H2O\mathrm{2\,C_6H_5CHO + CH_3COCH_3 \xrightarrow{} C_{17}H_{14}O + 2\,H_2O}

Molar masses: Benzaldehyde = 106 g mol−1106\,\mathrm{g\,mol^{-1}} Dibenzalacetone = 234 g mol−1234\,\mathrm{g\,mol^{-1}}

Moles of benzaldehyde used =5.3106=0.05 mol= \dfrac{5.3}{106} = 0.05\,\mathrm{mol}

From stoichiometry, 2 mol2\,\mathrm{mol} benzaldehyde →1 mol\rightarrow 1\,\mathrm{mol} product

Moles of product (theoretical) =0.052=0.025 mol= \dfrac{0.05}{2} = 0.025\,\mathrm{mol}

Mass of product (theoretical) =0.025×234=5.85 g= 0.025 \times 234 = 5.85\,\mathrm{g}

Percentage yield =3.515.85×100=60%= \dfrac{3.51}{5.85} \times 100 = 60\%

Thus, the percentage yield is 60%60\%.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
29 January, Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic
Chemical Equations, Stoichiometry and Limiting Reagent

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