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A 20.0 cm3 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 cm3. A further contraction of 14.0 cm3 occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.

Amit Saxena , 10 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra

Last Activity: 10 Years ago

(i) He does not react with oxygen.
(ii) KOH absorbs only CO2.
NOTE : When the mixture of CO, CH4 and He gases (20 ml) are exploded by an electric discharge with excess of O2, He gas remain as such and the other reactions involved are :
CO(g) + 1/2 O2(g) → CO2(g) …..(i)
CH4(g) + 2O2(g) → CO2(g) + 2H2O (l) …..(ii)
Let the volumes of CO and CH4 to be ‘a’ ml and ‘b’ ml in the mixture then
Volume of He gas = [20 – (a + b)] ml
For the initial contraction of 13 ml
For the initial contraction of 13 ml,
Volume of left hand side in the above reactions – 13 = volume
or right hand side.
∴ [20 – (a +b)] + (a + 1/2 a) + (b + 2b) – 13
= [20 – (a + 2b)] + a + b [neglect the volume of H2O (l)]
(Since for gases, volume ∝ no. of moles)
∴ 1/2 a + 2b = 13 or a + 4b = 26 …….(iv)
NOTE THIS STEP : The CO2 produced above in reactions (ii) & (iii), (a + b) ml, reacts with KOH sol for a further contraction of 14 ml.
CO2(g) + 2KOH(l) → K2 CO3 (l) + H2O(l)
(a + b) ml
∴ a + b = 14 ….(v)
Solving (iv) & (v) we get, a = 10ml &b = 4 ml
∴ CH4 = 4/20 * 100 = 20%, CO = 10/20 * 100 = 50%
& He = 100 – (20 + 50) = 30%

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