90 ml of gaseous mixture consisting of a gaseous organic compound A and just sufficient amount of oxygen required for complete combustion , yealding on burning 40 ml of CO2 , 60 ml of water vapour and 20 ml of N2. All volume are measured at same temp. and pressure ,compound A contains only carbon ,hydrogen and nitrogen . Now find the volume of the compound A initially present .
And also the molecular formula of compound A .
Answers
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
C
is it correct??
