M ethyl benzoate is prepared by the reaction between benzoic acid and methanol, according to the equation

$\underset{\mathrm{Benzoic} \mathrm{acid} }{{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}}+\underset{\mathrm{Methanol}}{{\mathrm{CH}}_3\mathrm{OH}}⟶\underset{\mathrm{Methyl} \mathrm{benzoate} }{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COOCH}}_3}+{\mathrm{H}}_2\mathrm{O}$

 In an experiment 24.4 gm 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 21.6 g. What is the percentage yield of product?