Methylation of 10 g of benzene gave 9.2 g o | vedclass

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(B) The chemical reaction for the methylation of benzene is: $C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl$. Molar mass of benzene $(C_6H_6)$ = $78 \

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$78$

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(B) The chemical reaction for the methylation of benzene is: $C_6H_6 + CH_3Cl ightarrow C_6H_5CH_3 + HCl$. Molar mass of benzene $(C_6H_6)$ = $78 \ g/mol$. Molar mass of toluene $(C_6H_5CH_3)$ = $92 \ g/mol$. Moles of benzene taken = $\frac{10 \ g}{78 \ g/mol} = 0.1282 \ mol$. According to the stoichiometry,$1 \ mol$ of benzene produces $1 \ mol$ of toluene. Theoretical yield of toluene (in grams) = $0.1282 \ mol 	imes 92 \ g/mol = 11.794 \ g$. Percentage yield = $\frac{	ext{Actual yield}}{	ext{Theoretical yield}} 	imes 100$. Percentage yield = $\frac{9.2 \ g}{11.794 \ g} 	imes 100 \approx 78 \ \%$.

Methylation of $10 \ g$ of benzene gave $9.2 \ g$ of toluene. Calculate the percentage yield of toluene $......$. (Nearest integer)

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