WHAT WOULD BE PERCENTAGE COMPOSITION BY VOLUME OF CO AND CH4 ,WHOSE 10 5 ml REQUIRES 9 ml OXYGEN FOR - Chemistry - States of Matter

Question

WHAT WOULD BE PERCENTAGE COMPOSITION BY VOLUME OF CO AND
CH4 ,WHOSE 10 5 ml REQUIRES 9 ml OXYGEN FOR COMPLETE
COMBUSTION?

CO=80% AND CH4=20%
CO=76 2% AND CH4=23 8%
CO=66% AND CH4=34%
CO=90% AND CH4 =10%

Arti Gujrati · Accepted Answer

Let us write two separate combustion reactions, one for CO and the
other for CH4 
reacting with O2
2CO(g)
+ O2 →2CO2
eq 
1
2mol 
1mol 2 mol
CH4 
(g) + 2O2 →CO2 
+ H2O(l)
eq 
2

1 mol 2 mol 
1mol zero (liquid)

Let us assume that out of 10 5 ml of mixture, the volume of CO = x
ml, and so the volume of CH4 
=(10 5-x)ml

From eq 
1,
2mol 
of CO produces 2 mol 
of CO2 
gas, so x mol 
of CO will produce x ml of CO2 
gas

Again 2 ml of CO requires 1 ml of O2,

so x ml of CO will require x/2 ml of O2

Further we have, from eq 
2,

1 mol 
of CH4 
produces 1 mol 
of CO2 
gas so, (10 5-x)ml of CH4 
will produce (10 5-x) ml of CO2

Again 1 mol 
of CH4 
requires 2 mol 
of O2 
gas so, (10 5-x)ml of CH4

requires 2(10 5-x)ml of O2 
gas

The volume of oxygen consumed by both reactions=x/2 + 2(10
5-x)=(42-3x)/2

We are provided with volume of oxygen consumed so,

(42-3x)/2
= 9 ml

42 - 3x 
= 18

Or, 3x 
= 42-18 = 24

Hence, x = 8 ml
So,
CO = 8 ml

and, CH4 
= 10 5-8 =2 5 ml
So,
%age composition of CO = (8/ 10 5) x 100 = 76 2%

%age composition of CH4 
= (2 5/ 10 5) x 100 = 23 8 %

WHAT WOULD BE PERCENTAGE COMPOSITION BY VOLUME OF CO AND CH4 ,WHOSE 10.5 ml REQUIRES 9 ml OXYGEN FOR COMPLETE COMBUSTION? CO=80% AND CH4=20% CO=76.2% AND CH4=23.8% CO=66% AND CH4=34% CO=90% AND CH4 =10%

WHAT WOULD BE PERCENTAGE COMPOSITION BY VOLUME OF CO AND CH₄ ,WHOSE 10.5 ml REQUIRES 9 ml OXYGEN FOR COMPLETE COMBUSTION?

CO=80% AND CH4=20%

CO=76.2% AND CH4=23.8%

CO=66% AND CH4=34%

CO=90% AND CH4 =10%