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%#