$20$kmo$\frac{l}{m}in$ nitrogen and $40$kmo$\frac{l}{m}in$ hydrogen are mixed with a recycle stream of hydrogen and nitrogen as shown below and fed into a reactor to synthesize ammonia. The exit stream from the reactor enters a scrubber. Water is added to the scrubber. All the ammonia is removed in a $15\%($mass$)$ water solution in the scrubber.

$2$zmo$\frac{l}{m}inN{H}_3,$ykmo$\frac{l}{m}in{N}_2$ and xkmo$\frac{l}{m}in{H}_2$ leave the reactor. Thus, $2$zmo$\frac{l}{m}inN{H}_3$ leaves the scrubber as a $15\%$ ammonia solution and ykmo$\frac{l}{m}in{N}_2$ and xkmo$\frac{l}{m}in{H}_2$ leave the top of the scrubber, where the stream is split and half is returned to the feed and half is purged.

$1.$Add the feed and recycle streams of $N_2$ and $H_2$ to find expressions for the streams entering the reactor.

$2.$ Let zmo$\frac{l}{m}in$ of the nitrogen, and $3$zmo$\frac{l}{m}in$ of the hydrogen streams entering the reactor be converted to form $2$zkmo$\frac{l}{m}in$ of ammonia. Write $y$ and $x$ in terms of $z.$ What is the maximum value of $z$ and what is the limiting reagent.

$3.$ The equilibrium in the reactor is in terms of concentration, not flow rates. But you can express flow rates of a species into concentration by dividing it with the total volume flow rate, $Q(\frac{m^3}{m}in).$ The equilibrium condition can be written as:

$K_{eq}=2(kmo\frac{l}{m^3}{)}^{-2}=\frac{[2\frac{z}{Q}{]}^2}{[\frac{x}{Q}{]}^3[\frac{y}{Q}]}$

Use the results from part $2$ to substitute for $y$ and $x$ in terms of $z$ and solve the equilibrium equation for

$Q=0\mathrm{.}2\frac{m^3}{m}in$ and $Q=1\frac{m^3}{m}in.$

$4.$ Find the amount of ammonia produced for both flow rates. Then calculate the water that must be added to the scrubber in each case $(k\frac{g}{m}in).$ How does an increase in Q affect the amount of ammonia that is formed?