Question 745de

Jun 9, 2017

Q-1

ns of combustion reactions are

${C}_{4} {H}_{8} + 6 {O}_{2} \to 4 C {O}_{2} + 4 {H}_{2} O$

${C}_{4} {H}_{10} + \frac{9}{2} {O}_{2} \to 4 C {O}_{2} + 5 {H}_{2} O$

Molar masses of

${C}_{4} {H}_{8} \to 56 g \text{/} m o l$

${C}_{4} {H}_{10} \to 58 g \text{/} m o l$

Let the number moles of the components in the mixture be

${C}_{4} {H}_{8} \to x m o l = 56 x g l$

${C}_{4} {H}_{10} \to y m o l = 58 y$

By the problem $56 x + 58 y = 2.86 \ldots \ldots \left[1\right]$

Using the balanced equation we can write the total amount of $C {O}_{2}$ produced

$4 x + 4 y$ mole

By the problem

$4 x + 4 y = \frac{8.8}{44} = 0.2$

$\implies x + y = \frac{0.2}{4} = 0.05 \ldots \ldots . \left[2\right]$

Multiplying  by 56 and subtracting the resulting equation from  we get

$2 y = 2.86 - 0.05 \times 56 = 0.06$

$\implies y = 0.03 m o l$

The mass of this component i.e.${C}_{4} {H}_{10}$ is $0.03 \times 58 g = 1.74 g$

So the percentage of ${C}_{4} {H}_{10}$ in the mixture

=1.74/2.86xx100%~~60.84%

So the percentage of ${C}_{4} {H}_{8}$ in the mixture

=(100-60.84)%=39.16%

Q-2

Let the formula of gaseous hydrocarbon be ${C}_{x} {H}_{y}$ and the balanced equation of the combustion reaction is

${C}_{x} {H}_{y} \left(g\right) + \left(x + \frac{y}{4}\right) {O}_{2} \left(g\right) \to x C {O}_{2} \left(g\right) + \frac{y}{2} {H}_{2} O \left(l\right)$

$1 m l \text{ "" "" "(x+y/4)ml" "" "" } x m l$

$V m l \text{ "" "" "V(x+y/4)ml" "" "" } V x m l$

Now 1st contraction

$= V + V \left(x + \frac{y}{4}\right) - V x = 2.5 V$

$\implies V + V \frac{y}{4} = 2.5 V$

$\implies 1 + \frac{y}{4} = 2.5$

$\implies y = \left(2.5 - 1\right) \times 4 = 6$

And 2nd contraction = volume of $C {O}_{2} \left(g\right)$ produced

$V x = 2 V$

$\implies x = 2$

So formula of hydrocarbon is ${C}_{2} {H}_{6}$