Question

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 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=(PV)/(RT)`

`n=(PV)/(RT)=((25atm)* (1100L))/((0.0821L.atm/mole.K)* (523K))`

`n=640.45moles`

For
`H_2` :

`PV=nRT\\\\n=(PV)/(RT)`

`n=(PV)/(RT)=((25atm)* (1550L))/((0.0821L.atm/mole.K)* (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* \text{Molar mass of }C_2H_6`

`\text{Mass of }C_2H_6=(640.45 mole)* (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}* 100=(14900g)/(19213.5g)* 100=77.55\%`

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