9.3 g of aniline is subjected to reaction wi | vedclass

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(A) The chemical reaction is: $C_6H_5NH_2 + (CH_3CO)_2O ightarrow C_6H_5NHCOCH_3 + CH_3COOH$Molar mass of aniline $(C_6H_5NH_2)$ = $(6 	imes 12) +

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$135$

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(A) The chemical reaction is: $C_6H_5NH_2 + (CH_3CO)_2O ightarrow C_6H_5NHCOCH_3 + CH_3COOH$Molar mass of aniline $(C_6H_5NH_2)$ = $(6 	imes 12) + (7 	imes 1) + 14 = 93 \ g \ mol^{-1}$.Molar mass of acetanilide $(C_6H_5NHCOCH_3)$ = $(8 	imes 12) + (9 	imes 1) + 14 + 16 = 135 \ g \ mol^{-1}$.Number of moles of aniline = $\frac{9.3 \ g}{93 \ g \ mol^{-1}} = 0.1 \ mol$.Since the reaction is $100 \%$ complete,$0.1 \ mol$ of aniline will produce $0.1 \ mol$ of acetanilide.Mass of acetanilide = $0.1 \ mol 	imes 135 \ g \ mol^{-1} = 13.5 \ g$.Given that the mass is $x 	imes 10^{-1} \ g$,we have $13.5 = x 	imes 10^{-1}$,which implies $x = 135$.

$9.3 \ g$ of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is $100 \%$ completed is $x \times 10^{-1} \ g$. Find the value of $x$. (Given molar mass in $g \ mol^{-1}: N: 14, O: 16, C: 12, H: 1$)

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