chem easy calculation ,easy pt

1.) A student performs an esterification reaction between benzoic acid and methanol to give methyl benzoate and water. The student starts with 1.14 moles of benzoic acid and 8.3 moles of methanol. At the end of the reaction 0.18 moles of benzoic acid are recovered. How much water is present at equilibrium?

C6H5COOH + CH3OH ⇌ C­6H5COOCH­3 + H2O

No. of moles of C­6H5COOCH­3 reacted = 1.14 - 0.18 = 0.96 mol
No. of moles of H2O formed = 0.96 mol


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2.) A student is performing an esterification reaction with benzoic acid and methanol to give methyl benzoate and water. The student starts with 0.97 moles of benzoic acid and 6.7 moles of methanol. At the end of the experiment 0.92 moles of methyl benzoate are isolated. Assuming no experimental loses, how much benzoic acid should be recovered?

C6H5COOH + CH3OH ⇌ C­6H5COOCH­3 + H2O

No. of moles of C­6H5COOCH­3 formed = 0.92 mol
No. of moles of C6H5COOH reacted = 0.92 mol
No. of moles of C6H5COOH recovered = 0.97 - 0.92 = 0.05 mol


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3.) If 12.31 grams of benzoic acid is allowed to react with 6.0 moles of methanol and 12.72 grams of methyl benzoate are isolated, what is the percent yield for this transformation?

C6H5COOH + CH3OH ⇌ C­6H5COOCH­3 + H2O
Molar mass of C6H5COOH = 12.01x7 + 1.008x6 + 16x2 = 122.12 g/mol
No. of moles of C6H5COOH added = 12.31/122.12 = 0.1008 mol < 6.0 mol

Hence, maximum yield in mol of C­6H5COOCH­3 = 0.1008 mol
Molar mass of C­6H5COOCH­3= 12.01x8 + 1.008x8 + 16x2 = 136.14
Maximum yield in g of C­6H5COOCH­3 = 136.14 x 0.1008 = 13.72 g
Percentage yield = (12.72/13.72) x 100% = 92.7%