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7nadin3 [17]
3 years ago
7

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

zoate. Calculate the theoretical yield and percent yield for this reaction.
Chemistry
1 answer:
Anit [1.1K]3 years ago
6 0

<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:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For benzoic acid:</u>

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol

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

\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}

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 = \frac{1}{1}\times 0.0295=0.108 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:

0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g

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

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%

Hence, the percent yield of the reaction is 32.34 %

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Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

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