Question

Ethene is converted to ethane by the reaction

Answer 1

Answer : The percent yield of the reaction is, 77.55 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1100 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1100L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=640.45moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 640.45 mole of $C_2H_4$ react with 640.45 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$.

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 640.45 mole of $C_2H_4$ react to give 640.45 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$.

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(640.45 mole)\times (30g/mole)=19213.5g$

The theoretical yield of $C_2H_6$ = 19213.5 g

The actual yield of $C_2H_6$ = 14.9 kg = 14900 g      (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14900g}{19213.5g}\times 100=77.55\%$

Therefore, the percent yield of the reaction is, 77.55 %