# Question 100b4

Jun 8, 2017

Molar masses of

${C}_{2} {H}_{2} \to 2 \times 12 + 2 \times 1 = 26 g \text{/} m o l$

${C}_{2} {H}_{4} \to 2 \times 12 + 4 \times 1 = 28 g \text{/} m o l$

$C {H}_{4} \to 1 \times 12 + 4 \times 1 = 16 g \text{/} m o l$

The mole ratio in the mixture

${C}_{2} {H}_{2} : {C}_{2} {H}_{4} : C {H}_{4} = 2 : 1 : 2$

The mass ratio in the mixture

${C}_{2} {H}_{2} : {C}_{2} {H}_{4} : C {H}_{4} = \left(2 \times 26\right) : \left(1 \times 28\right) : \left(2 \times 16\right) = 13 : 7 : 8$

The masses of the components in the gas mixture of 1g

${C}_{2} {H}_{2} \to \frac{13}{28} g \to \frac{13}{28} \times \frac{1}{26} m o l = \frac{1}{56} m o l$

${C}_{2} {H}_{4} \to \frac{7}{28} g \to \frac{7}{28} \times \frac{1}{28} m o l = \frac{1}{112} m o l$

$C {H}_{4} \to \frac{8}{28} g \to \frac{8}{28} \times \frac{1}{16} m o l = \frac{1}{56} m o l$

Balanced equation of the reactions

${C}_{2} {H}_{2} \left(g\right) + \frac{5}{2} {O}_{2} \left(g\right) \to 2 C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$

${C}_{2} {H}_{4} \left(g\right) + 3 {O}_{2} \left(g\right) \to 2 C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

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

So the number of moles of $\textcolor{red}{{O}_{2} \left(g\right)}$ required for complete burning of component gases are

${C}_{2} {H}_{2} \to \frac{1}{56} \times \frac{5}{2} = \frac{5}{112} m o l$

${C}_{2} {H}_{4} \to \frac{1}{112} \times 3 = \frac{3}{112} m o l$

$C {H}_{4} \to \frac{1}{56} \times 2 = \frac{4}{112} m o l$

So total amount of $\textcolor{red}{{O}_{2} \left(g\right)}$ required for complete burning of 1 g gas mixture is

${O}_{2} \to \frac{5 + 3 + 4}{112} m o l = \frac{12}{112} m o l$

As the ratio of ${O}_{2} : {N}_{2}$ by volume in air is (20%):(80%)=1:4#

The mole ratio of ${O}_{2} : {N}_{2}$ in air will be $= 1 : 4$

So in air $\frac{12}{112} m o l$ ${O}_{2}$ will be with $\frac{12}{112} \times 4 m o l$ ${N}_{2}$

Hence total mass of air required for complete burning is

$\frac{12}{112} m o l$ ${O}_{2} +$ $\frac{12}{112} \times 4 m o l$ ${N}_{2}$

$= \frac{12 \times 32 + 48 \times 28}{112} g = \frac{1728}{112} g \to$ option A