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A reaction was performed in which 3.0 g of benzoic acid was reacted with excess methanol...

A reaction was performed in which 3.0 g of benzoic acid was reacted with excess methanol to make 2.9 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.

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Answer #1

Solution:

Given that weight of Benzoic acid=3.0g

Number of moles of benzoic acid =weight/molecular weight=3.0/122.12=0.024566moles

weight of Methyl benzoate=2.9g

number of moles of methyl benzoate=weight/molecular weight=2.9/136.14=0.02130

from the above reaction one mole of benzoic acid reacts with ethanol to produce one mole of methyl benzoate.

then 0.024566moles of benzoic acid shoud produce 0.024566moles of methyl benzoate.

weight of 0.024566moles of methyl benzoate =3.34g

theoritical yield=3.34g

%percent yield=(Actual yield/theoritical yield)*100=(2.9/3.34)*100=86.826%

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