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80 ml of oxygen is added to 50 ml of a mixture of H2, C2H2 and CO after which the total mixture is burnt completely. The volume of the cooled mixture after combustion is 65ml. This is further reduced to 15 ml by treatment with KOH solution.calulate the volume of each gas in the mixture.

80 ml of oxygen is added to 50 ml of a mixture of H2, C2H2 and CO after which the total mixture is burnt completely. The volume of the cooled mixture after combustion is 65ml. This is further reduced to 15 ml by treatment with KOH solution.calulate the volume of each gas in the mixture.

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Grade:12

1 Answers

Dhananjai
13 Points
2 years ago
First by considering the balanced reactions;
c2h2+co+h2=50ml
Let;
c2h2=x,co=y,h2=50-x-y
 
c2h2+5/2o2---->2co2+h2o
co+1/2o2---->co2
h2+1/2o2---->h2o
 
Since 65ml of mixture was obtained after cooling and it reduced to 15ml after reacting with KOH solution, It means there was (65-15)=50 ml of co2 in mixture;
Co2 is produced only by C2H2 and CO;
So according to stoichometric coefficients;
2x+y=50----(eq.1)
Amount of o2 reacted is 80ml(initial) minus 15ml(leftover)
80-15=65ml
From stoichometry;
(5/2)*x+(1/2)*y+(50-x-y)/2=65
5x+y+50-x-y=130
By solving the above equation,we get;
4x=80
X=20ml
So c2h2=20ml
From (eq.1)
y=10ml
h2=50-x-y=50-20-10=20ml
So;
h2=20ml
c2h2=20ml
co=10ml
So correct option is (1).
 
 
 
 
 

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