Respuesta :
Answer : The percent yield of the reaction is, 76.34 %
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] = 1050 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 (1050L)}{(0.0821L.atm/mole.K)\times (523K)}[/tex]
[tex]n=611.34moles[/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, 611.34 mole of [tex]C_2H_4[/tex] react with 611.34 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, 611.34 mole of [tex]C_2H_4[/tex] react to give 611.34 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=(611.34mole)\times (30g/mole)=18340.2g[/tex]
The theoretical yield of [tex]C_2H_6[/tex] = 18340.2 g
The actual yield of [tex]C_2H_6[/tex] = 14.0 kg = 14000 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{14000g}{18340.2g}\times 100=76.34\%[/tex]
Therefore, the percent yield of the reaction is, 76.34 %
