From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ______ g

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Solution

Q.
From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ______ $\times 10^{- 1}$ g. [2024]

Ans.
(95)

Number of moles of aniline $(n_{C_{6} H_{5} {NH}_{2}}) = \frac{Given mass}{Molar mass} = \frac{6.55}{93} mol$

Maximum 0.07 mol of acetanilide can be obtained from 0.07 mol of aniline.

Maximum amount of acetanilide = number of moles × molar

mass of acetanilide = $\frac{6.55}{93} \times 135 g = 9.5 g = 95 \times 10^{- 1} g$

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