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

Answer:

a) 17,20$\frac{g}{mol}$

b) 87,57%

c) 0,7259$\frac{g}{L}$

Explanation:

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

3,9%×30,07$\frac{g}{mol}$ + 1,1%×44,1$\frac{g}{mol}$ +1,1%×44,01$\frac{g}{mol}$ + 93,9%×16,04$\frac{g}{mol}$ = 17,20$\frac{g}{mol}$

b) The weight fraction of CH₄ is:

$\frac{0,939.16,04}{17,20}$×100 = 87,57%

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

$\frac{P.MW}{R.T}$ = δ

Where:

P is pressure: 148 kPa

MW is molecular weight: 17,20$\frac{g}{mol}$

R is gas constant: 8,31$\frac{L.kPa}{K.mol}$

T is temperature: 422K.

And δ is density. Replacing you will obtain:

δ = 0,7258$\frac{g}{mol}$

I hope it helps!