200 mL of a mixture of 50% H_2,40% CH_4 an | vedclass

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(C) Initial composition of $200 \ mL$ mixture:$H_2 = 50\% 	ext{ of } 200 = 100 \ mL$$CH_4 = 40\% 	ext{ of } 200 = 80 \ mL$$CO_2 = 10\% 	ext{ of } 2

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$100$

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(C) Initial composition of $200 \ mL$ mixture:$H_2 = 50\% 	ext{ of } 200 = 100 \ mL$$CH_4 = 40\% 	ext{ of } 200 = 80 \ mL$$CO_2 = 10\% 	ext{ of } 200 = 20 \ mL$Combustion reactions:$2H_2(g) + O_2(g) 	o 2H_2O(l)$$100 \ mL \ H_2$ consumes $50 \ mL \ O_2$ and produces $0 \ mL \ CO_2$.$CH_4(g) + 2O_2(g) 	o CO_2(g) + 2H_2O(l)$$80 \ mL \ CH_4$ consumes $160 \ mL \ O_2$ and produces $80 \ mL \ CO_2$.Total $CO_2$ produced = $80 \ mL$ (from $CH_4$) + $20 \ mL$ (initial $CO_2$) = $100 \ mL$.When passed through $aq. KOH$,$CO_2$ is absorbed:$CO_2(g) + 2KOH(aq) 	o K_2CO_3(aq) + H_2O(l)$Volume contraction = Volume of $CO_2$ absorbed = $100 \ mL$.

$200 \ mL$ of a mixture of $50\% \ H_2$,$40\% \ CH_4$ and $10\% \ CO_2$ is exploded with excess of oxygen in an eudiometry tube. The resultant mixture is cooled to room temperature and is passed through $aq. KOH$. Find volume contraction after passing through $KOH$ ................. $mL$

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