Question

60 ml of sample containing co, ch4 and h2 requires 70ml of o2 for complete combustion of 40ml when the mixture eas treated with koh what is the combustion of co and ch4 respectively in original mixture?

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Answer 1

Answer:

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Answer 2

Answer:

CO Volume = 2 ml

Explanation:

$2co + o2$⇒$2co2$

$2h2 + o2$⇒$h2o$

( 60 - × ) $( 60 - x )/2$

after that,

oxygen used = ×/2 + 60 - $x$/2 = 30 ml

residual oxygen = 40 - 30 = 10 ml

vol of $co2$ + residual oxygen = $x$ + 10 = 30

or

co volume = 2 ml

h2 volume = 40 ml