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

here first we will convert lbs to grams ,

10 lbs = 4535.92 gms ,

30 lbs= 13607.8 gms,

and final mixture = 30 lbs= 18143.7 gms,

we have 20 mole % methane in 10 lbs means in 4535.92 gms*20 %=907.184 gms ,

20 mol % ethane = 4535.92 gms *20% =907.184 gms ,

60 mol % propane = 4535.92*60%= 2721.55 gms ,

in 30 lbs we have ,13607.8*20 % ethane= 2721.56 gms ,

13607.8*35 % propane *35 %= 4762.73 gms,

13607.8 *25 % n-butane = 3401.95 gms , (2721.56 +4762.73+3401.95=10886.24),

13607.8-10886.24=2721.56 gms (other may be water)

now we can calculate the over all composition of mixture as,

methane = 907.184/18143.7=0.05*100=5.0 %,

ethane= 907.84 +2721.56=3629.4/18143.7=0.200*100=20.0%,

propane = 2721.55 +4762.73=7484.28/18143.7=0.4125*100=41.25 %,

butane = 3401.95 /18143.7=0.1875*100=18.75 %

water= 2721.56/18143.7=0.15*100=15 %

total= 5%+20% +41.25%+18.75%+15%=100%