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A 20 mL mixture of CO, CH4, and Helium (...

A 20 mL mixture of CO, `CH_4`, and Helium (He) gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13 mL. A further contraction of 14 mL occurs when the residual gas is treated wityh KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.

Text Solution

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Let `CO="aml "CH_4=bml`
Helium `=[20-(a+b)]ml`
(i). `underset(a (a)/(2) a)CO+(1)/(2)O_2rarrCO_2`
(ii) `underset(b 2b b)(CH_4+2)O_2rarrCO_2+2H_2O`
(iii) `Heimplies[20-(a+b)]`
(helium does not react, it will remain same.) Initial volume
`(a+(a)/(2)+b+2b)+[20-(a+b)]`
Final volume:
`a+b+[20-(a+b)]`
Contraction `=13ml`
Initial voue `=`final volume
`a+(a)/(2)+b+2b+[20-(a+b)]-13`
`=(a+b)+[20-(a+b)]`
`implies(a+(a)/(2)+b+2b)-13=a+b`
`impliesa+b=14`
Solving, `a=10,b=4`
Percentage of `CO=(10)/(20)xx100=50%`
percentage of `CH_4=(4)/(20)xx100=20%`
percentage of of He `=100-(50+20)=30%`
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