# If 1.87 g of acetic acid(CH3CO2H) reacts with 2.31 g of isopentyl alcohol(C5H12O) to give 2.96 g...

## Question:

If 1.87 g of acetic acid(CH3CO2H) reacts with 2.31 g of isopentyl alcohol(C5H12O) to give 2.96 g of isopentyl acetate(C7H14O2), what is the percent yield of the reaction?

## Limiting Reagent

Limiting Reagent is the one which is present in limiting or very less amount on the reactants side.

In technical terms, we can say that the reagent whose number of moles are less as compare to another reagent that is present in the main chemical reaction, that reactant is known as limiting reagent.

Step 1 Find the limiting reagent first

Calculation of the number of moles of acetic acid

{eq}\dfrac{1.87\ g}{60.052\ g/mol}\ =\ 0.031\ moles\\ {/eq}

Calculation of the number of moles of isopentyl alcohol

{eq}\dfrac{2.31\ g}{88.148 \ g/mol}\ =\ 0.026\ moles\\ {/eq}

The number of moles of isopentyl alcohol that are reacting is less than the number of moles of acetic acid that are reacting.

Therefore, isopentyl alcohol is the limiting reagent

88.148 g/mol isopentyl alcohol produces 130.19 g/mol isopentyl acetate

1 g isopentyl alcohol produces {eq}\dfrac{130.19\ g/mol }{88.148\ g/mol}\ {/eq} isopentyl acetate

2.31 g isopentyl alcohol produces {eq}\dfrac{130.19\ g/mol \times 2.31\ g}{88.148\ g/mol}\ =\ 3.4\ g {/eq} isopentyl acetate

Theoretical yield = 3.4 g

Experimental yield = 2.96

Step 2 Calculation of percentage yield

% Yield = {eq}\dfrac{Experimental\ yield }{ Theoretical\ yield}\times 100\\ \dfrac{2.96 }{ 3.4}\times 100\\ {/eq}

The percentage yield is 87%.