Question

A sample of liquefied natural, LNG, from Alaska has the following molar composition:3.9% C2H6, 1.1% C3H8, 1.1% CO2, and the rest CH4. Determine (a) The average molecular weight of the LNG mixture. (b) The weight fraction of CH4 in the mixture. (c) The LNG is heated to 422 K and 148 kPa, and vaporizes completely. Estimate the density of the gas mixture under these conditions.

Answer 1

a) 17,20
`(g)/(mol)`

b) 87,57%

c) 0,7259
`(g)/(L)`

Step-by-step explanation:

a) With molar composition of LNG the average molecular weight is:

3,9%×30,07
`(g)/(mol)` + 1,1%×44,1
`(g)/(mol)` +1,1%×44,01
`(g)/(mol)` + 93,9%×16,04
`(g)/(mol)` = 17,20
`(g)/(mol)`

b) The weight fraction of CH₄ is:

`(0,939.16,04)/(17,20)`×100 = 87,57%

c) It is possible to use ideal gas law thus:

`(P.MW)/(R.T)` = δ

Where:

P is pressure: 148 kPa

MW is molecular weight: 17,20
`(g)/(mol)`

R is gas constant: 8,31
`(L.kPa)/(K.mol)`

T is temperature: 422K.

And δ is density. Replacing you will obtain:

δ = 0,7258
`(g)/(mol)`

I hope it helps!