<u>Answer:</u> The percent yield of the reaction is 32.34 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of benzoic acid = 3.6 g
Molar mass of benzoic acid = 122.12 g/mol
Putting values in equation 1, we get:

The chemical equation for the reaction of benzoic acid and methanol is:

By Stoichiometry of the reaction
1 mole of benzoic acid produces 1 mole of methyl benzoate
So, 0.0295 moles of benzoic acid will produce =
moles of methyl benzoate
- Now, calculating the mass of methyl benzoate from equation 1, we get:
Molar mass of methyl benzoate = 136.15 g/mol
Moles of methyl benzoate = 0.0295 moles
Putting values in equation 1, we get:

- To calculate the percentage yield of methyl benzoate, we use the equation:

Experimental yield of methyl benzoate = 1.3 g
Theoretical yield of methyl benzoate = 4.02 g
Putting values in above equation, we get:

Hence, the percent yield of the reaction is 32.34 %