In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is _________g. (Nearest integer)
The reaction involved is the Claisen-Schmidt condensation: 2C6H5CHO+CH3COCH3NaOHC6H5CH=CHCOCH=CHC6H5+H2O. Step 1: Reaction Stoichiometry 2 moles of benzaldehyde (C6H5CHO) react with 1 mole of acetone (CH3COCH3) to form 1 mole of dibenzalacetone (C6H5CH=CHCOCH=CHC6H5). Given that 87 g of acetone is used, the molar mass of acetone is: Molar mass of CH3COCH3=12+(1×3)+12+16+1=58g/mol. Moles of acetone=Molar mass of acetoneMass of acetone=5887≈1.5moles. Step 2: Calculate moles of benzaldehyde required From the reaction stoichiometry, 2 moles of benzaldehyde are required for every 1 mole of acetone. Thus, the moles of benzaldehyde needed are: Moles of benzaldehyde=2×Moles of acetone=2×1.5=3moles. Step 3: Calculate the mass of benzaldehyde required The molar mass of benzaldehyde (C6H5CHO) is: Molar mass of benzaldehyde=(6×12)+(5×1)+12+16=106g/mol. Mass of benzaldehyde=Moles of benzaldehyde×Molar mass of benzaldehyde=3×106=318g. Step 4: Verify with product formation The reaction produces 1 mole of dibenzalacetone for every 2 moles of benzaldehyde. For 1.5 moles of acetone, 1.5 moles of dibenzalacetone are formed, which corresponds to 351 g (given). Final Answer: 318 g.
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