)200 ml of a gaseous mixture containing co and co2 and 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 ration of VCo2: VCO: V N2 in the original mixture.
Answers
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 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
CHEMISTRY
200 ml of gaseous mixture containing CO, CO
2
and N
2
on complete combustion in just sufficient amount of O
2
showed a contraction of 40 ml . When the resulting gases were passed through KOH solution it reduces by 50 ml, then calculate the volume of V
CO
:V
CO
2
:V
N
2
in the original mixture.
Share
Study later
ANSWER
Given, Volume of CO, CO
2
,N
2
=200 ml
On combustion, CO
2
remains intact and nitrogen does not react at low temperatures.
2CO +O
2
→2CO
2
x x/2 x
Total Volume = 200 +
2
x
Resulting mixture = N
2
= y ml
CO
2
=200−y ml
Contradiction = Final - Initial = 200 +
2
x
−200=
2
x
40=
2
x
Hence x=80 ml
When mixture passes through KOH, only CO
2
reacts.
As volume reduces by 50% ,
200−y=100 ml
y=100 ml
Hence CO=80 ml, y=100 ml and CO
2
= 20 ml
V
CO
:V
CO
2
:V
N
2
=4:1:5