Question #1d13b

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Answer 1 · 2017-06-11T12:08:03

Q-1

Let the volume of $CH_4$ in 20ml gas mixture be $x$ mL and the volume of hdrocarbon of acetylene series be $(y)$ mL.

So $x+y=20......[1]$

Let us also assume that the formula of the gas of actylene series be $C_nH_(2n-2)$ ,the alkyne with one $CequivC$ bond

Now balanced equation of combustion reaction of $CH_4$

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

$xmL" "" "" "2xmL" "" "xmL$

Here first cotraction $=(x+2x-x)=2x$ mL

Now balanced equation of combustion reaction of $C_nH_(2n-2)$

$C_nH_(2n-2)(g)+((3n-1)/2)O_2(g)->nCO_2(g)+(n-1)H_2O(l)$

$ymL" "" "" "((3n-1)/2)ymL" "" "nymL$

Here first cotraction

$=(y+((3n-1)/2)y-ny)$ mL

By the equation total first contraction

$2x+(y+((3n-1)/2)y-ny)$ mL

By the problem this first contraction $=(20+100-80)=40mL$

Hence

$2x+(y+((3n-1)/2)y-ny)=40.....[2]$

Combining [1] and [2] we get

$40-2y+y+((3n-1)/2)y-ny)=40$

$=>-y+((3n-1)/2)y-ny=0$

$=>-1+(3n-1)/2-n=0$

$=>-2+3n-1-2n=0$

$=>n=3.....[3]$

So the molecular formula the hydrocarbon acetylene series

$C_3H_4->"propyne"$

2nd contraction will be the total volume of $CO_2(g)$ produced and by the promlem it is $40$ mL

So

$x+ny=40$

$=>x+3y=40.....[4]$

By [1] and [4] we get

$x=10mL and y=10mL$

So percentage of each gas by volume will be $10/(10+10)xx100%=50%$