Please show all steps clearly. Problem # 8: A gas mixture has the following comp
ID: 883867 • Letter: P
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Please show all steps clearly.
Problem # 8: A gas mixture has the following composition on a molar basis: Calculate: (a) Mixture composition (mass fraction) of 100 kg-mole of mixture. (b) Average Molecular Weight of the mixture expressed in lbm/lb-mole and g/g-mole. (c) How many grams of C2H6 are in 50 lb-mole of mixture? (d) How many kg-mole of CH4 are in 20 kg of mixture? (e) How many grams of CO2 are in the mixture when there are 15 g-mole of NO2? (f) Mixture mole fraction (CO2 and NO2 free basis)Explanation / Answer
a) the total mole fraction of a gas mixture will be equal to 1
So xCH4 + xC2H6 + xCO2 + xNO2 = 1
xNO2 = 1-(0.1 + 0.2 + 0.3 ) = 0.4
Mass = Mole fraction X total moles X molecular weight
mCH4 = 0.1 X 100 X 16 = 160 Kg
mC2H6 = 0.2 X 100 X 30 = 600 Kg
mCO2 = 0.3 X 100 X 44 = 1320 Kg
mNO2 = 0.4 x 100 X 46 = 1840 Kg
So Total mass = 1840 +1320 + 600 + 160 = 3920 Kg
Mass fraction = mass of gas / total mass
Mass fraction CH4 = 160 / 3920 = 0.0408
Mass fraction C2H6 = 600 / 3920 = 0.1531
Mass fraction CO2 = 1320 / 3920 = 0.3367
Mass fraction NO2 = 1840 / 3920 = 0.4694
b) Average molecular weight
= mole fraction of methane x molecular weight of methane + mole fraction of ethane X molecular weight of ethane + mole fraction of CO2 X molecular weight of CO2 + molecular weight of NO2 X mole fraction of NO2 = 39.2
c) mC2H6 = 50lb mixture X 0.2 lb moles X 30 lb C2H6 X 454 grams = 136200 grams
d) kg moles of methane = amount of mixture in kg X mass fraction of methane / molecular weight
= 20 x 0.0408 / 16 = 0.051
e) When 0.4 g mole of NO2 is present then 0.3 g mole of CO2 present
so when 1gmole of NO2 then 0.3 / 0.4 g mole
So 15 g mole of NO2 then 0.3 X 15 / 0.4 = 11.25 g mole
so amount = moles X molecular weight of CO2 = 11.25 X 44 = 495 grams
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