occurs through two reactions:

$2C{H}_4+3O_{2}2CO+4H_{2}O.$

$C{H}_4+2O_{2}C{O}_2+2H_{2}O.$

Given that the feed stream $($stream $1)$ has $52$ mole $\%N_2,$ with $C{H}_4$ and $O_2$ at stoichiometric proportion with respect to reaction $[1].$ The feed enters into the reactor and then to the separator which separates some of the products with the unreacted material. The unreacted material is recycled to the feed stream. The separator is not $100\%$ efficient and thus some unreacted material go along with the product streams and some products go into the recycle stream. There are two product streams, one contains water $($stream $4)$ and the other stream contains $C{O}_2,CO,H_{2}O,C{H}_4$ and $O_2($stream $3).$ Since the feed stream contains nitrogen gas which does not react, therefore a purge stream is required to remove the nitrogen gas.

Using the data given below calculate the flow rate of the stream that contain water only $(n_{4+20}).$

Overall pass conversion $C{H}_4$ is $76\%.$

Flow rate of stream $1$ is $100$kmo$\frac{l}{h}r$

Extent of reaction $[1]$ is $3\mathrm{.}5$kmo$\frac{l}{h}r$

$\backslash$table$[[$Stream$,1,3,4,6],[O_2(\%),,3\mathrm{.}7,,5\mathrm{.}3],[CO(\%),,30\mathrm{.}8,,3\mathrm{.}7],[C{O}_2(\%),43\mathrm{.}9,,3\mathrm{.}9,],[H_{2}O(\%),,18\mathrm{.}6,100,24\mathrm{.}5],[C{H}_4(\%),,3\mathrm{.}0,,4\mathrm{.}3],[N_2(\%),,,,58],[,52,,,]]$

Purge $(6)$