Question #745de

1 Answer
Jun 9, 2017

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#