Question

10.0 lb-moles of ternary Mixture #1 (20 mol.% methane, 20 mol.% ethane, and 60 mol.% propane) is mixed with 30.0 lbs of ternary Mixture #2 (20 wt.% methane, 20 wt.% ethane, 35 wt.% propane and 25 wt.% n-butane), determine the overall composition of the final mixture.

Answer 1

moless in 10 lb moles of ternary mixture : Methane = 20*10/100 =2 moles, ethane = 2 moles and propane = 6 moles

for the mixture-2,

Molar masses : CH4= 16, C2H6= 30 and propnae(C3H8)=44, C4H10(butane)= 58

Mass of components in mixture-2 ( lb): CH4= 30*20/100= 6, C2H6= 6 , propane = 35*30/100= 10.5, butane = 30*25/100 = 7.5

Moles = mass/molar mass : CH4= 6/16 = 0.375, ethane= 6/30= 0.2, propane= 10.5/44 =.24 and butane= 7.5/58 = 0.13

Total moles of Mixture-2 =0.375+0.2+0.24+0.13= 0.94

when mixture-1 and mixture -2 are mixed, moles : 10+0.94= 10.94 lb moles

moles in mixed sample : CH4= 2+0.375=2.375, C2H6= 2+0.2= 2.2m Propane= 6+0.24 =6.24 and butane =0.13

composition ( mole %):   100*(moles of given component/totla moles)

Composition ( Mole %): CH4=100*2.375/10.94=21.71 % , C2H6= 100*2.2/10.94= 20.11, propane= 100*6.24/10.94= 57.04, Butane= 100*0.13/10.94=1.19