natural gas contains 84% CH4, 10% CH6, 3% C3H8, 3% N2...?

Question

in a series of rxns all C atmos are converted to gas C4H6 with efficiency 80%. how much C4H6 is prepared from 1000 ltr of natural gas at 10 atm pressure and 400K temp?

thnx for pointin the error

its C2H6 ¬ CH6. N2 is nitrogen gas

its C2H6 and not CH6

Answer 1

Can you write again the correct formulas because CH6 does not exist and also does N2 mean nitrogen or a compound ?
1000 L of natural gas contains
840 L of CH4
moles CH4 = 10 atm x 840 / 0.0821 x 400 = 255.8
2 CH4 >> C2H6 + H2
if the efficiency would be 100% we get 255.8 / 2 = 127.9 moles of C2H6
80 : 100 = x : 127.9
x = 102.3 moles C2H6

in 1000 L we have also 100 L of C2H6
moles C2H6 = 100 x 10 / 0.0821 x 400 = 30.45

in 1000 L we have 30 L of C3H8
moles C3H8 = 30 x 10 / 0.0821 x 400 = 9.13
2 C3H8 + H2 >> 3C2H6
the ratio between C3H8 and C2H6 is 2 : 3
2 : 3 = 9.13 : x
x = 13.7 moles C2H6 ( efficiency 100%)
80 : 100 = x : 13.7
x = 11.0 moles C2H6

total moles of C2H6 = 102.3 + 30.45 + 11.0 = 143.75

Molar mass = 30 g/mol
143.75 mol x 30 g/mol = 4312.5 g = 4.3125 g
I hope this will help you

Answer 2

You seem to have problems with the actual product C2H6 (ethane). 1000 L of natural gas will provide 840 L of methane. With 80 pct efficiency, 672 L are available to be produced.
The reaction is 2 CH4 -> C2H6 + H2 ?, so each 2 moles of CH4 produces 1 mole of ethane. Getting back to our problem, the 672 L are converted to moles with PV=nRT, where P=10 atm, T=400 K and R= 22.4/273. The mole of product are computed above.

Answer 3

Data required.