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.