Question

. KMnO4 oxidizes ethyl benzene to benzoic acid. If this reaction generally results in a 40 % experimental yield, how many grams of ethyl benzene are required to obtain 244 g of benzoic acid

Answer 1

Answer:

530.835 g

Explanation:

First we convert 244 g of benzoic acid (C₇H₆O₂) to moles, using its molar mass:

• 244 g benzoic acid ÷ 122.12 g/mol = 2.00 moles benzoic acid

Theoretically, one mol of ethyl benzene would produce one mol of benzoic acid. But the experimental yield tells us that one mol of ethyl benzene will produce only 0.4 moles of benzoic acid.

With the above information in mind we convert 2.00 moles of benzoic acid into moles of ethyl benzene:

• 2.00 moles benzoic acid * $\frac{1molEthylBenzene}{0.4molBenzoicAcid}$ = 5.00 moles ethyl benzene

Finally we convert moles of ethyl benzene (C₈H₁₀) into grams, using its molar mass:

• 5.00 moles ethyl benzene * 106.167 g/mol = 530.835 g ethyl benzene