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# 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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