# Question 1d13b

Jun 11, 2017

Q-1

Let the volume of $C {H}_{4}$ in 20ml gas mixture be $x$ mL and the volume of hdrocarbon of acetylene series be $\left(y\right)$ mL.

So $x + y = 20. \ldots . . \left[1\right]$

Let us also assume that the formula of the gas of actylene series be ${C}_{n} {H}_{2 n - 2}$,the alkyne with one $C \equiv C$ bond

Now balanced equation of combustion reaction of $C {H}_{4}$

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

$x m L \text{ "" "" "2xmL" "" } x m L$

Here first cotraction$= \left(x + 2 x - x\right) = 2 x$mL

Now balanced equation of combustion reaction of ${C}_{n} {H}_{2 n - 2}$

${C}_{n} {H}_{2 n - 2} \left(g\right) + \left(\frac{3 n - 1}{2}\right) {O}_{2} \left(g\right) \to n C {O}_{2} \left(g\right) + \left(n - 1\right) {H}_{2} O \left(l\right)$

$y m L \text{ "" "" "((3n-1)/2)ymL" "" } n y m L$

Here first cotraction

$= \left(y + \left(\frac{3 n - 1}{2}\right) y - n y\right)$mL

By the equation total first contraction

$2 x + \left(y + \left(\frac{3 n - 1}{2}\right) y - n y\right)$mL

By the problem this first contraction$= \left(20 + 100 - 80\right) = 40 m L$

Hence

$2 x + \left(y + \left(\frac{3 n - 1}{2}\right) y - n y\right) = 40. \ldots . \left[2\right]$

Combining [1] and [2] we get

40-2y+y+((3n-1)/2)y-ny)=40

$\implies - y + \left(\frac{3 n - 1}{2}\right) y - n y = 0$

$\implies - 1 + \frac{3 n - 1}{2} - n = 0$

$\implies - 2 + 3 n - 1 - 2 n = 0$

$\implies n = 3. \ldots . \left[3\right]$

So the molecular formula the hydrocarbon acetylene series

${C}_{3} {H}_{4} \to \text{propyne}$

2nd contraction will be the total volume of $C {O}_{2} \left(g\right)$ produced and by the promlem it is $40$ mL

So

$x + n y = 40$

$\implies x + 3 y = 40. \ldots . \left[4\right]$

By [1] and [4] we get

$x = 10 m L \mathmr{and} y = 10 m L$

So percentage of each gas by volume will be 10/(10+10)xx100%=50%#