Q-1
ns of combustion reactions are
#C_4H_8+6O_2->4CO_2+4H_2O#
#C_4H_10+9/2O_2->4CO_2+5H_2O#
Molar masses of
#C_4H_8->56g"/"mol#
#C_4H_10->58g"/"mol#
Let the number moles of the components in the mixture be
#C_4H_8->xmol =56xgl#
#C_4H_10->ymol=58y#
By the problem #56x+58y=2.86......[1]#
Using the balanced equation we can write the total amount of #CO_2# produced
#4x+4y# mole#
By the problem
#4x+4y=8.8/44=0.2#
#=>x+y=0.2/4=0.05.......[2]#
Multiplying [2] by 56 and subtracting the resulting equation from [1] we get
#2y=2.86-0.05xx56=0.06#
#=>y=0.03mol#
The mass of this component i.e.#C_4H_10# is #0.03xx58g=1.74g#
So the percentage of #C_4H_10# in the mixture
#=1.74/2.86xx100%~~60.84%#
So the percentage of #C_4H_8# in the mixture
#=(100-60.84)%=39.16%#
Q-2
Let the formula of gaseous hydrocarbon be #C_xH_y# and the balanced equation of the combustion reaction is
#C_xH_y(g)+(x+y/4)O_2(g)->xCO_2(g) +y/2H_2O(l)#
# 1ml" "" "" "(x+y/4)ml" "" "" "xml#
# Vml" "" "" "V(x+y/4)ml" "" "" "Vxml#
Now 1st contraction
#=V+V(x+y/4)-Vx=2.5V#
#=>V+Vy/4=2.5V#
#=>1+y/4=2.5#
#=>y=(2.5-1)xx4=6#
And 2nd contraction = volume of #CO_2(g)# produced
#Vx=2V#
#=>x=2#
So formula of hydrocarbon is #C_2H_6#
