The amount (in mol) of bromoform (CHBr_{3}) p | vedclass

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(A) The reaction of acetone with bromine in the presence of aqueous $NaOH$ is the haloform reaction. The balanced chemical equation for the reaction o

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$\frac{1}{3}$

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(A) The reaction of acetone with bromine in the presence of aqueous $NaOH$ is the haloform reaction. The balanced chemical equation for the reaction of acetone with bromine is: $CH_{3}COCH_{3} + 3Br_{2} + 4NaOH ightarrow CHBr_{3} + CH_{3}COONa + 3NaBr + 3H_{2}O$ From the stoichiometry of the reaction,$3 \ mol$ of $Br_{2}$ are required to produce $1 \ mol$ of $CHBr_{3}$. Given that $1.0 \ mol$ of bromine is used,the amount of $CHBr_{3}$ produced is: $	ext{Amount of } CHBr_{3} = \frac{1}{3} 	imes 	ext{Amount of } Br_{2} = \frac{1}{3} 	imes 1.0 \ mol = \frac{1}{3} \ mol$.

The amount (in $mol$) of bromoform $(CHBr_{3})$ produced when $1.0 \ mol$ of acetone reacts completely with $1.0 \ mol$ of bromine in the presence of aqueous $NaOH$ is $......$

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