Chemistry, 04.12.2019 00:31 Aydenj9613

Assume that you react 100 mg of benzoic acid with 10 ml of methanol and 10 microliters of sulfuric acid to produce methyl benzoate. write a balance chemical equation for this reaction. determine the limiting reagent and calculate a theoretical yield of both the ester and water. if you isolate 75 mg of methyl benzoate, what is the actual yield of the reaction?

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Answer from: bryce99

Explanation:

Balanced equation for the reaction between benzoic acid and metanol is as follows.

$C_{6}H_{5}COOH + CH_{3}OH \overset{H_{2}SO_{4}}{\rightarrow} C_{6}H_{5}COOCH_{3}$

Since, volume of $H_{2}SO_{4}$ is very small so, that is catalytic amount of $H_{2}SO_{4}$ which is used.

As mass of benzoic acid is 100 mg. Hence, moles of benzoic acid are calculated as follows.

No. of moles of benzoic acid = $\frac{mass}{molar mass}$

= $\frac{100 mg}{122.12 g/mol}$

= 0.818 mmol

And, mass of methanol = volume × density

= $10 ml \times 0.792 g/mol$

= 7.92 g

Now, number of moles of methanol is as follows.

No. of moles of methanol = $\frac{mass}{molar mass}$

= $\frac{ 7.92 g}{32.04 g/mol}$

= 0.024 mol

= 24.7 mmol

As number of moles of benzoic acid are smaller than the number of moles of methanol. Hence, benzoic acid is the limiting reagent.

As per the balanced equation, 1 mole of benzoic acid produces 1 mole of methyl benzoate.

Hence, 0.818 mmol of benzoic acid would produce 0.818 mmol of methyl benzoate. Therefore, theoretical yield of methyl benzoate is as follows.

Theoretical yield of methyl benzoate = $0.818 mmol \times 136.15 g/mol$

= 111.48 mg

or, = 111.5 mg (approx)

Now, we will calculate the percent yield of the reaction as follows.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

= $\frac{75 mg}{111.5 mg} \times 100$

= 67.27%

Therefore, we can conclude that the actual yield of the reaction is 67.27%.

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Assume that you react 100 mg of benzoic acid with 10 ml of methanol and 10 microliters of sulfuric a...

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