From 6.55 g of aniline,the maximum amount of | vedclass

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Question

(D) The reaction of aniline $(C_6H_5NH_2)$ with acetic anhydride forms acetanilide $(C_6H_5NHCOCH_3)$.Molar mass of aniline $(C_6H_5NH_2)$ = $93 \ g/m

Vedclass · Accepted Answer

$95$

Vedclass · Answer

(D) The reaction of aniline $(C_6H_5NH_2)$ with acetic anhydride forms acetanilide $(C_6H_5NHCOCH_3)$.Molar mass of aniline $(C_6H_5NH_2)$ = $93 \ g/mol$.Molar mass of acetanilide $(C_6H_5NHCOCH_3)$ = $135 \ g/mol$.According to the stoichiometry,$1 \ mol$ of aniline produces $1 \ mol$ of acetanilide.$93 \ g$ of aniline produces $135 \ g$ of acetanilide.Therefore,$6.55 \ g$ of aniline produces $\frac{135}{93} 	imes 6.55 \ g = 9.508 \ g \approx 9.5 \ g$.Converting to the required format: $9.5 \ g = 95 	imes 10^{-1} \ g$.

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

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