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Question
A 20.0 cc mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13.0 cc. A further contraction of 14.0 cc occurs when the residual gas is treated with KOH solution. The composition of the gaseous mixture in terms of volume percentage will be :
A 20.0 cc mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13.0 cc. A further contraction of 14.0 cc occurs when the residual gas is treated with KOH solution. The composition of the gaseous mixture in terms of volume percentage will be :
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Solution
The correct option is D CO:50% , CH4:20% , He:30%
20 cc mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of oxygen. Only CO and CH4 will react with oxygen while He is inert towards it.
The reaction of CO with O2 is as follows:
CO+12O2→CO2Let a cc be the volume of CO and b cc be the volume of CH4 in the mixture.Now, a volume of CO will react with a2 volume of O2 to produce a volume of CO2.
The reaction of CH4 with O2 is as follows:
CH4+2O2→CO2+2H2O
b volume of CH4 will react with 2b volumes of O2 to produce b volume of CO2.
14 cc of CO2 is absorbed by KOH.
Therefore, a+b=14 ----- (1)
Volume of He in the mixture which does not react with O2 is 20−14=6 cc
Volume percentage of He in the mixture =6×10020=30%
Initial volume of the mixture and excess oxygen =20+V, where V cc is the volume of excess oxygen added.
Final volume of the mixture after reaction =[a+b] (Volume of CO2)+[V−a2−2b] (Volume of unreacted O2) +6 (unreacted He) =V+a2−b+6
Change in volume =20+V−V−a2+b−6=13
Thus, we get a2−b=1 ----- (2)
Solving (1) and (2), we get a=10 and b=4.
Thus, 10 volumes of CO and 4 volumes of CH4 is present in the mixture.
Volume percentage of CO=10×10020=50%
Volume percentage of CH4=4×10020=20%
Hence, the composition of the gaseous mixture is CO:50%, CH4:20% and He:30%.
20 cc mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of oxygen. Only CO and CH4 will react with oxygen while He is inert towards it.
The reaction of CO with O2 is as follows:
CO+12O2→CO2
CO+12O2→CO2
Let a cc be the volume of CO and b cc be the volume of CH4 in the mixture.
Now, a volume of CO will react with a2 volume of O2 to produce a volume of CO2.
The reaction of CH4 with O2 is as follows:
CH4+2O2→CO2+2H2O
b volume of CH4 will react with 2b volumes of O2 to produce b volume of CO2.
CH4+2O2→CO2+2H2O
b volume of CH4 will react with 2b volumes of O2 to produce b volume of CO2.
14 cc of CO2 is absorbed by KOH.
Therefore, a+b=14 ----- (1)
Volume of He in the mixture which does not react with O2 is 20−14=6 cc
Volume percentage of He in the mixture =6×10020=30%
Initial volume of the mixture and excess oxygen =20+V, where V cc is the volume of excess oxygen added.
Final volume of the mixture after reaction =[a+b] (Volume of CO2)+[V−a2−2b] (Volume of unreacted O2) +6 (unreacted He) =V+a2−b+6
Change in volume =20+V−V−a2+b−6=13
Thus, we get a2−b=1 ----- (2)
Solving (1) and (2), we get a=10 and b=4.
Thus, 10 volumes of CO and 4 volumes of CH4 is present in the mixture.
Volume percentage of CO=10×10020=50%
Volume percentage of CH4=4×10020=20%
Hence, the composition of the gaseous mixture is CO:50%, CH4:20% and He:30%.
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