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%#
