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Question

200 ml of a gaseous mixture containing CO,CO2,N2 ON COMPLETE COMBUSTION IN JUST SUFFICIENT AMOUNT OF O2 SHOWED CONTRACTION OF 40 ML.WHEN THE RESULTING GASES WERE PASSED THROUGH KOH SOLUTION IT REDUCES BY 50% THEN CALCULATE THE VOLUME RATIO OF CO2:CO:N2 IN ORIGINAL MIXTURE

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Solution

Volume of mixture of CO, CO2 and N2 = 200
ml

Volume of CO = x ml

Volume of N2 = y ml

Volume of CO2 = 200 - x - y ml


On combustion, CO2 remains as it is. Nitrogen burns only at very
high temperatures. At low temperatures it does not form oxides.

2 CO + O2 ==> 2 CO2

x ml x/2 ml
x ml


In the input the amount of O2 present = x/2 ml

Total volume of gas mixture + O2 = 200 + x/2 ml


Resulting mixture: total : 200 ml as:

N2: y ml
CO2: x + (200 - x - y) = 200 - y ml


Contraction (reduction) in volume of gases is

40 ml = 200 + x/2 - 200 = x/2

x = volume of CO in the mixture = 80
ml


Now the mixture of CO2 + N2 is passed through base KOH. All CO2 reacts and N2
is not reactive with K OH.

2 K OH + CO2 ==> K2 CO3 +
H2 O


So volume of gas mixture reduces by 50% of 200 ml ie., 100 ml.

=> 200 - y = 100 ml

=> y = 100 ml


So we had 80 ml of CO, 100 ml of N2, and 20 ml of CO2 in the mixture initially.

Ratio: of volumes of CO2 : CO : N2 = 1 : 4 : 5



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