Question

20 ml mixture of CO and C2H4 was exploded with 50 ml of O2.The volume after explosion was 45 ml on shaking with NaOH solution only 15 ml O2 was left behind ,what were the volume of CO and C2H4 in mixture?

Answer 1

CH4 & C2H4 only reacts with O2 to produce CO2 & H2O.
The CO2 present in the original mixture do not react and remains as such
let the volume of CH4=x ml
volume of C2H4=yml,
Therefore:volume of CO2=(10-x-y)ml
contraction in volume due to explosion=17ml
and contraction of volume after treatment with KOH=14ml=volume of CO2 produced.
the combustion reaction are;CH4(g)+2O2(g)............>CO2(g)+2H2O(l).........(i)
xml 2xml xml
C2H4(g)+3O2......>2CO2(g)+2H2O(l)............(ii)
yml 3yml 2yml
contraction in volume in reaction.................(i)
y+2x-x=2xml
contraction in volume in reaction...............(ii)
y+3y-2y=2yml
Total contraction in volume=2(x+y)ml
Now
2(x+y)=17...........(ii)
& total volume of CO2 produced in reaction..........(i) &(ii),=x+2yml
The total volume of CO2 present in the mixture=x+2y+10-x=(y+10)ml
the volume of CO2 is absorbed in KOH,
THEREFORE;
y+10=14 or y=4ml
 substituting this value in equation(iii)we get:
x=4.5ml
10-x-y=10-4.5-4=1.5ml
volume of CH4 in the mixture =4.5ml
volume of C2H4 in the mixture=4ml
volume of CO2 in the mixture=1.5ml

Answer 2

Explanation:

Let the mixture contain y ml of C2H4 and x ml

COC2H4 + 3O2 → 2CO2 + 2H2O

y 2y

CO + 12O2 → CO2

x x

NaOH absorbs the CO2 produced in the reaction

Total amount of CO2 produced = x+2y =(45−15) ml

and we know that, x+y=20 ml

Therefore, (20−y)+2y=30

or, 20+y=30

or, y=10 ml

Therefore, x=10 ml