A process for methanol synthesis is shown in the flow diagram

below. The chemical reactions involved are $($a$)$ Main former

reaction: CH$4+2$H$2$O $=$ CO$2+4$H$2($b$)$ Reformed Side Reaction: CH$4+$

H$2$O $=$ CO $+3$H$2($c$)$ CO Converter reaction: $2$CO $+$ O$2=2$ CO$2($d$)$

Methanol synthesis reaction: CO$2+3$H$2=$ CH$3$OH $+$ H$2$O Ten percent

excess steam, based on the "main former reaction" is fed to the

reformer, and conversion of methane is $100\%,$ with a $90\%$ yield of

CO$2.$ Conversion in the methanol reactor is $50\%$ on one pass through

the reactor. A stoichiometric quantity of oxygen is fed to the CO

converter, and the CO is completely converted to CO$2.$ Additional

makeup CO$2$ is then introduced to established a $3-$to$-1$ ratio of H$2$

to CO$2$ in the feed stream to the methanol reactor. The methanol

reactor effluent is cooled to condense all of the methanol and

water, with the noncondensable gases recycled to the methanol

reactor feed. The H$2/$ CO$2$ ratio in the recycle stream is $3$ to $1.$

Because the methane feed contains $1\%$ nitrogen as an impurity, a

portion of the recycle stream must be purged as shown in the figure

below to prevent the accumulation of nitrogen in the system. The

pruge stream analyzes $5\%$ nitrogen. On the basis of $100$ mol of

methane feed $($including the N$2)$ calculate: a$.$ The number of moles

of H$2$ are lost in the purge b$.$ The number of moles of makeup CO$2$

required c$.$ The recycle$-$to$-$purge ratio in moles per mole d$.$ The

methanol solution $($in kilograms$)$ of what strength $($weight percent$)$

produced.