Question

An ideal gas is mixture of $C_2{H}_6$ and $C_2{H}_4$ occupies $28$l at $1$ atm $8273$ k. The mixture reacts completely with $128$ g of $O_2$ to produce $C{O}_2$ pf water vapour. Mole fraction of $C_2{H}_6$ in the mixture will be:

• A. 0.4
• B. 0.6
• C. 0.2
• D. 0.8

Answer 1

The correct option is A 0.4

Given $P=1atm$

$V=28l$

$T=8273$

$128g$ of $O_2$ produce $C{O}_2$

mixture of $C_2{H}_6$ & $C_2{H}_4$

$PV=nRT$

$1\times 28=n\times 0.082\times 8273$

total moles $=0.04127$

$C_2{H}_6+C_2{H}_4=0.04127$

$a+b=\mathrm{0.04127...}\left(1\right)$

$C_2{H}_6+\frac{7}{2}{O}_2\to 2C{O}_2+3H_{2}O$

$C_2{H}_4+3O_2\to 2C_2+2H_{2}O$

$\frac{7a}{2}+3b=\frac{128}{32}...\left(2\right)$

find value of a & b from $e{q}^n\left(1\right)$ & $\left(2\right)$

mole fraction of $C_2{H}_6=\frac{a}{totalmoles}$