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

first find moles of all copound in 1 st mixture

moles of methane = (10 / 100) x 20 % = 2.0 lb moles

moles of ethane = (10 /100) x 20 % = 2.0 lb moles

moles of propane = (10 / 100) x 60% = 6.0 lb moles

find moles of all copound in 2nd mixture

moles of methane = (30 / 100) x 20 % = 6.0 lb moles

moles of ethane = (30 /100) x 20 % = 6.0 lb moles

moles of propane = (30 / 100) x 35% = 10.5 lb moles

moles of n-butane = (30 / 100 )x 25% = 7.5 lb moles

overall composition

8.0 lb moles of methane, 8.0 lb moles of ethane, 16.5 lb of propane, 7.5 moles of n-butane

mixture contain 40.0 lb moles ( 20 mole % methane, 20 mole% ethane, 41.25 moles% propane, 18.75 moles% n-butane)