Question

Toulene, C6H5CH3 is oxidized by air under carefully controlled conditions to benzoic acid, C6H5CO2H, which is used to prepare the food perservative sodium benzoate, C6H5CO2Na. What is the percent yield os a reaction that converts 1000 kg of toulene to 1.21 kg of benzoic acid
2C6 H5 CH3 + 3O2 ? 2C6 H5 CO2 H + 2H2O

Answer 1

Answer : The percent yield of the reaction is, 0.0913 %

Explanation : Given,

Mass of $C_6H_5CH_3$ = 1000 kg  = 1000000 g

Molar mass of $C_6H_5CH_3$ = 98.14 g/mole

Molar mass of $C_6H_5COOH$ = 122.12 g/mole

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

$\text{Moles of }C_6H_5CH_3=\frac{\text{Mass of }C_6H_5CH_3}{\text{Molar mass of }C_6H_5CH_3}=\frac{1000000g}{92.14g/mole}=10853.05moles$

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

The balanced chemical reaction is,

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

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, 10853.05 moles of $C_6H_5CH_3$ react to give 10853.05 moles of $C_6H_5COOH$

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

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

$\text{Mass of }C_6H_5COOH=(10853.05mole)\times (122.12g/mole)=1325374.466g$

Theoretical yield of $C_6H_5COOH$ = 1325374.466 g

Actual yield of $C_6H_5COOH$ = 1.21 kg = 1210 g

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

$\%\text{ yield of }C_6H_5COOH=\frac{\text{Actual yield of }C_6H_5COOH}{\text{Theoretical yield of }C_6H_5COOH}\times 100=\frac{1210g}{1325374.466g}\times 100=0.0913\%$

Therefore, the percent yield of the reaction is, 0.0913 %