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Ethene is converted to ethane by the reaction

C2H4 flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1100. L/min. Hydrogen at 25.0 atm and 250.°C flows into the reactor at a flow rate of 1550. L/min. If 14.9 kg is collected per minute, what is the percent yield of the reaction?

Answer :

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

Explanation : Given,

Pressure of [tex]C_2H_4[/tex] and [tex]H_2[/tex] = 25.0 atm

Temperature of [tex]C_2H_4[/tex] and [tex]H_2[/tex] = [tex]250^oC=273+250=523K[/tex]

Volume of [tex]C_2H_4[/tex] = 1100 L per min

Volume of [tex]H_2[/tex] = 1550 L per min

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

Molar mass of [tex]C_2H_6[/tex] = 30 g/mole

First we have to calculate the moles of [tex]C_2H_4[/tex] and [tex]H_2[/tex] by using ideal gas equation.

For [tex]C_2H_4[/tex] :

[tex]PV=nRT\\\\n=\frac{PV}{RT}[/tex]

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

[tex]n=640.45moles[/tex]

For [tex]H_2[/tex] :

[tex]PV=nRT\\\\n=\frac{PV}{RT}[/tex]

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

[tex]n=902.46moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]C_2H_4+H_2\rightarrow C_2H_6[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]C_2H_4[/tex] react with 1 mole of [tex]H_2[/tex]

So, 640.45 mole of [tex]C_2H_4[/tex] react with 640.45 mole of [tex]H_2[/tex]

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

Now we have to calculate the moles of [tex]C_2H_6[/tex].

As, 1 mole of [tex]C_2H_4[/tex] react to give 1 mole of [tex]C_2H_6[/tex]

As, 640.45 mole of [tex]C_2H_4[/tex] react to give 640.45 mole of [tex]C_2H_6[/tex]

Now we have to calculate the mass of [tex]C_2H_6[/tex].

[tex]\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6[/tex]

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

The theoretical yield of [tex]C_2H_6[/tex] = 19213.5 g

The actual yield of [tex]C_2H_6[/tex] = 14.9 kg = 14900 g      (1 kg = 1000 g)

Now we have to calculate the percent yield of [tex]C_2H_6[/tex]

[tex]\%\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\%[/tex]

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

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