Question

Toluene, C6H5CH3, is oxidized by air under carefully controlled conditions to benzoic acid, C6H5CO2H, which is used to prepare the food preservative sodium benzoate, C6H5CO2Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?

Answer 1

Answer : The percent yield of the reaction is, 91.32 %

Explanation :  Given,

Mass of $C_6H_5CH_3$ = 1 Kg = 1000 g

Molar mass of $C_6H_5CH_3$ = 92.14 g/mole

Molar mass of $C_6H_5COOH$ = 122.12 g/mole

First we have to calculate the moles of $C_6H_5CH_3$.

$\text{Moles of }C_6H_5CH_3=\frac{\text{Mass of }C_6H_5CH_3}{\text{Molar mass of }C_6H_5CH_3}=\frac{1000g}{92.14g/mole}=10.85mole$

Now we have to calculate the moles of $C_6H_5COOH$.

The balanced chemical reaction will be,

$2C_6H_5CH_3+3O_2\rightarrow 2C_6H_5COOH+2H_2O$

From the balanced reaction, we conclude that

As, 2 moles of $C_6H_5CH_3$ react to give 2 moles of $C_6H_5COOH$

So, 10.85 moles of $C_6H_5CH_3$ react to give 10.85 moles of $C_6H_5COOH$

Now we have to calculate the mass of $C_6H_5COOH$

$\text{Mass of }C_6H_5COOH=\text{Moles of }C_6H_5COOH\times \text{Molar mass of }C_6H_5COOH$

$\text{Mass of }C_6H_5COOH=(10.85mole)\times (122.12g/mole)=1325.002g$

The theoretical yield of $C_6H_5COOH$  = 1325.002 g

The actual yield of $C_6H_5COOH$  = 1.21 Kg = 1210 g

Now we have to calculate the percent yield of $C_6H_5COOH$

$\%\text{ yield of }C_6H_5COOH=\frac{\text{Actual yield of }C_6H_5COOH}{\text{Theoretical yield of }C_6H_5COOH}\times 100=\frac{1210g}{1325.002g}\times 100=91.32\%$

Therefore, the percent yield of the reaction is, 91.32 %