A mixture of $C{H}_4$ & $O_2$ is used as an optimal fuel if $O_2$ is present in thrice the amount required theoretically for combustion of $C{H}_4$. Calculate number of effusions steps required to convert a mixture containing $1$ part of $C{H}_4$ in 193 parts mixture (parts by volume). If calorific value (heat evolved when 1 mole is burnt) of $C{H}_4$ is 100 cal/mole & if after each effusion 80% of $C{H}_4$ is collected. Find out what initial mole of each gas in initial mixture required for producing 1000 cal of energy after processing.
[Given $(0.9{)}^5=0.6$]

The correct option is C $10Steps,27.78molC{H}_4,5333.3mol{O}_2$

$C{H}_4+2O_2\to C{O}_2+H_{2}O$

3 times $O_2$ theoretical is required for optimal fuel

For optimal fuel $\frac{n_{C{H}_4}}{n_{O_2}}=\frac{1}{2\times 3}=\frac{1}{6}$

We know,

$\frac{r_{C{H}_4}}{r_{O_2}}=\frac{n_{C{H}_4}}{n_{O_2}}\times \sqrt{\frac{M_{O_2}}{M_{C{H}_4}}}$

$\frac{1}{6}=\frac{1}{192}\times (\sqrt{2}{)}^n$; $\frac{192}{6}=(\sqrt{2}{)}^n$

$32=(\sqrt{2}{)}^n$

n= 10 steps

Let initial moles be $n_A,n_B$

Afer 1 effusion

$\frac{n_A}{n_b}=\frac{n_{Ai}}{n_{Bi}}\times \sqrt{2}$

And 90% 0f $n_A$ is remove

$\frac{n_A}{n_O}=\frac{n_{Ai}}{n_{Bi}}\times \sqrt{2}\times 0.9$

After 10 steps

We know

1 mole $C{H}_4$ produces 100 - fours

to produces

$\frac{1}{3}=\frac{n_{Ai}}{n_{Bi}}\times \left(\sqrt{2}x\right)$

$\frac{1}{6}=\frac{n_{Ai}}{n_{Bi}}\times 32x$

Also o produce 1000 cal,10 moles of $C{H}_4$ is begin

$10=n_{C{H}_4}\times {0.9}^{10}$

$n_{C{H}_4}=27.78$ mole

MOles of $O_2=2778\times 3\times 36\times 32=5333.33$ moles