7.8 g of a compound having molecular formula | vedclass

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(A) Molar mass of benzene $= 78 \ g/mol$. Mole of benzene $= \frac{7.8 \ g}{78 \ g/mol} = 0.1 \ mol$. The reaction is $C_6H_6 + CH_3COCl \xrightarrow{

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$70$

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(A) Molar mass of benzene $= 78 \ g/mol$. Mole of benzene $= \frac{7.8 \ g}{78 \ g/mol} = 0.1 \ mol$. The reaction is $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_8H_8O + HCl$. According to the stoichiometry,$1 \ mol$ of benzene produces $1 \ mol$ of $C_8H_8O$. Therefore,$0.1 \ mol$ of benzene should theoretically produce $0.1 \ mol$ of $C_8H_8O$. Molar mass of $C_8H_8O = (8 	imes 12) + (8 	imes 1) + (1 	imes 16) = 96 + 8 + 16 = 120 \ g/mol$. Theoretical yield $= 0.1 \ mol 	imes 120 \ g/mol = 12 \ g$. Percentage yield $= \frac{	ext{Experimental yield}}{	ext{Theoretical yield}} 	imes 100 = \frac{8.4 \ g}{12 \ g} 	imes 100 = 70\%$.

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