Question #36044

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Answer 1 · 2016-10-16T04:51:18

The alkane is $CH_4$

General equation for combustion reaction of an alkane.

$C_xH_y(g)+(x+y/4)O_2->xCO_2(g)+y/2H_2O(g)$

By the condition of the question

$1+x+y/4=x+y/2$

$=>y/2-y/4=1$

$=>y/4=1$

$=>y=4$

Again the hydrocarbon being an alkane $y =2x+2$

$4=2x+2$

$=>2x=2$

$=>x=1$

So the alkane is $CH_4$ and equation becomes

$CH_4(g)+2O_2(g)->CO_2(g)+2H_2O(g)$

The presence of 3 moles gas both in reactant and product side in the balanced equation satisfies our answer.