Methyl benzoate is prepared by the reaction between benzoic acid and methanol, according to the equation C6H5COOH + CH3OH → C6H5COOCH3 + H2O Benzoic acid Methanol Methyl benzoate In an experiment 24.4 g of benzoic acid were reacted with 70.0 mL of CH3OH. The density of CH3OH is 0.79 g mL–1. The methyl benzoate produced had a mass of 6.8 g. What was the percentage yield of product ?

Engineering

Chemistry

Chemical Equation and Stoichiometric Calculation

Question

Methyl benzoate is prepared by the reaction between benzoic acid and methanol, according to the equation
C₆H₅COOH + CH₃OH → C₆H₅COOCH₃+ H₂O

Benzoic acid Methanol Methyl benzoate

In an experiment 24.4 g of benzoic acid were reacted with 70.0 mL of CH₃OH. The density of CH₃OH is 0.79 g mL^–1. The methyl benzoate produced had a mass of 6.8 g. What was the percentage yield of product ?

25 %

20 %

50 %

23 %

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CH₃COOH = 55.3 gm
C₆H₅COOH + CH₃OH → C₆H₅COOCH₃ + H₂O
$\frac{22 .4}{122}$ = 0.2 mol $\frac{55 .3}{32}$ = 1.73 mol $\frac{6 .8}{136}$ = 0.5
% yield = × 100 = 25 %