9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product ‘P’. The mass of product ‘P’ obtained is 26.4 g. The percentage yield is

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Solution

Q.
9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product ‘P’. The mass of product ‘P’ obtained is 26.4 g. The percentage yield is ______ %. [2024]

Ans.
(80)

$n_{C_{6} H_{5} N H_{2}} = \frac{W_{C_{6} H_{5} N H_{2}}}{M_{C_{6} H_{5} N H_{2}}} = \frac{9.3 g}{93 {gmol}^{- 1}} = 0.1 mol$

$n_{C_{6} H_{3} B r_{3} N H_{2}} = \frac{W_{C_{6} H_{3} B r_{3} N H_{2}}}{M_{C_{6} H_{3} B r_{3} N H_{2}}} = \frac{26.4 g}{330 {gmol}^{- 1}} = 0.08 mol$

If the reaction was 100% complete, the moles of $C_{6} H_{3} {Br}_{3} {NH}_{2}$ would be 0.1 mol.

$Percentage yield of reaction = \frac{Amount of product formed}{Amount of product expected for 100% yield} \times 100$

$= \frac{0.08}{0.1} \times 100 = 80 %$

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