Respuesta :
Answer : The percent yield of the reaction is, 91.32 %
Explanation : Given,
Mass of [tex]C_6H_5CH_3[/tex] = 1 Kg = 1000 g
Molar mass of [tex]C_6H_5CH_3[/tex] = 92.14 g/mole
Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mole
First we have to calculate the moles of [tex]C_6H_5CH_3[/tex].
[tex]\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[/tex]
Now we have to calculate the moles of [tex]C_6H_5COOH[/tex].
The balanced chemical reaction will be,
[tex]2C_6H_5CH_3+3O_2\rightarrow 2C_6H_5COOH+2H_2O[/tex]
From the balanced reaction, we conclude that
As, 2 moles of [tex]C_6H_5CH_3[/tex] react to give 2 moles of [tex]C_6H_5COOH[/tex]
So, 10.85 moles of [tex]C_6H_5CH_3[/tex] react to give 10.85 moles of [tex]C_6H_5COOH[/tex]
Now we have to calculate the mass of [tex]C_6H_5COOH[/tex]
[tex]\text{Mass of }C_6H_5COOH=\text{Moles of }C_6H_5COOH\times \text{Molar mass of }C_6H_5COOH[/tex]
[tex]\text{Mass of }C_6H_5COOH=(10.85mole)\times (122.12g/mole)=1325.002g[/tex]
The theoretical yield of [tex]C_6H_5COOH[/tex] = 1325.002 g
The actual yield of [tex]C_6H_5COOH[/tex] = 1.21 Kg = 1210 g
Now we have to calculate the percent yield of [tex]C_6H_5COOH[/tex]
[tex]\%\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\%[/tex]
Therefore, the percent yield of the reaction is, 91.32 %
