3 is the easiest. The others are the same kind but are a little more complicated. This will get you stated on #3.
CH4 + 2O2 ==> CO2 + H2O
2C2H6 + 7O2 ==> 4CO2 + 6H2O
If O2 is 90 cc initially and excess O2 was 32 cc, then 90-32 = 58 cc O2 used.
Let x = volume CH4
and (24-x) = volume C2H6
volume O2 used for CH4 is x*(2 mol O2/1 mol CH4) = 2x
volume O2 used for C2H6 is (24-x)*7/2 = 168-7x/2.
The sum of O2 used by CH4 and C2H6 is
2x + 168-7x/2 = 58
Solve for x and I get about 17cc O2 used for CH4 and about 7 cc for O2 used for combustion of C2H6.
Note that these are approximations. You need to go through the math and obtain better accuracy. Then use cc O2 used for each to calculate volume CH4 and volume C2H6 and from there you can find percentage.
1) 30cm3 of methane and ethane mixed with 100cm3 of O2 at RTP. Explosion occurs, on cooling 61.5cm3 remained. Find the percentage by volume of each gas in the mixture
2)32cm3 of a mix of CO2, CH4 and H2 mix with 50cm3 of O2 and exploded after cooling to RTP, volume noted is reduced by 22cm3 when exposed to KOH solution, 16cm3 of O2 is in excess. Calculate the composition of the mixture.
3) 24cm3 of a mixture of CH$ and C2H6 exploded with 90cm3 of O2 after cooling to RTP, volume was noted which was decreased by 32cm3 when treated with KOH solution. Calculate the composition of the mixture.
4) 195cm3 of hydrogen, carbon monoxide and nitrogen mixed with 150cm3 of oxygen exploded. Final volume after cooling is 165cm3. Final volume after absorbtion with KOH is 75cm3. Find the percentage by volume of each of the gases present.
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