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Grade: 11
        A 20.0 cm3 mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of further 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.
5 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							Sol. The reaction involved in the explosion process is :
CO(g) + 1/2 O2 (g) → CO2 (g)
x mL x/2 mL x mL
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l)
y mL 2 y mL y mL
The first step volume contraction can be calculated as :
(x +  x/2  + y + 2 y)-(x + y) = 13
⇒ x + 4 y = 26
The second volume contraction is due to absorption of CO2.
Hence,   x + y = 14   …. (ii)
Now, solving equations (i) and (ii),
X = 10 mL, y = 4 mL and volume of He = 20 – 14 = 6 mL
⇒ Vol % of CO = 10/20 × 100 = 50%
Vol of CH4¬ = 4/20 × 100 = 20%
Vol % of He = 30%

						
5 years ago
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