Question

C2H4 flows into a catalytic reactor at 25.0 atm and 300.°C with a flow rate of 956.7 L/min. Hydrogen at 25.0 atm and 300.°C flows into the reactor at a flow rate of 1500. L/min. If 11.0 kg of C2H6 are collected per minute, what is the percent yield of the reaction?

Answer 1

$PV =nRT$ hence,$n = PV /RT$

First convert all units into SI units

P=25*101325Pa, T = 573K, V=0.9567M3 and R =8.314j/mol/k

$C2H4+H2=C2H6$

To get the number of moles of ethene, we find the value of n.

N=2533125*0.9567/8.314*573

N=508.70moles but theoretical molar mass of ethene =30.07g/mol

Hence theoretical yield of ethene =508.70*30.07 =15296.609g which is 15.30kg

%Yield=$Actual yield/theoretical yield*100$

%Yield =11/15.30*100=71.90%