Navjyot Kalra
Last Activity: 10 Years ago
Let the volume of ethane in mixture = x litre
∴ Volume of ethane = (40 - x) litre
Combustion reactions f ethane and ethane are :
(i) C2H6(g) + 3 1/2 O2 (g) → 2CO2 (g) + 3H2O(l)
or 2C2H6(g) → 4CO2(g) + 6H2O(l)
(ii) C2H4(g) + 3O2(g) → 2CO2 (g) + 2H2O(l)
Volume of O2 required for complete combustion of ethane
= 7x/2 [For x litres]
Volume of O2 required for complete combustion of ethane
= (40 - x) * 3 [For (40 - x)L]
∴ Total volume of O2 required = 7x/2 + (40 - x)3 l
Calculation of number of moles (n),
P = 1 atm, V = 7x/2 + (40 - x)3 l ; R = 0.082 l atm K-1 mol-1;
T = 400 K
Since n = PV/RT = 1*[7x/2 +(40 - x)3]/0.082 *400 = 7x +(40 -x)6/2 *0.082 *400
Mass of n moles of O2 [7x +(40 -x)6 2 * 0.082 * 400] * 32 = 130
or 130 = [7x +6 240 – 6x/65.6] * 32
=> 8528 = 32x + 240 * 32 => 32x = 848 => or x = 848/32 = 26.5
Hence mole fraction (%) of ethane = 26.5/40 * 100 = 66.25%
Mole fraction (%) of ethane = 33.75%
