You are an engineer that have been put in charge for reactor cultivation of bacteria cells $(C_{4\mathrm{.}4}{H}_{7\mathrm{.}3}{O}_{1\mathrm{.}2}{N}_{0\mathrm{.}86})$ from hexadecane $(C_{16}{H}_{34})($Figure $1).$ The bacteria will be used for the production of COVID$-19$ vaccine. The reaction equation describing growth is:

$C_{16}{H}_{34}+{16\mathrm{.}280}_2+1\mathrm{.}43N{H}_{3}1\mathrm{.}65C_{4\mathrm{.}4}{H}_{7\mathrm{.}3}{O}_{1\mathrm{.}2}{N}_{0\mathrm{.}86}+8\mathrm{.}74C{O}_2+13\mathrm{.}11H_{2}O$

rigure $1$

$($a$)$ If the medium contains $3\%(\frac{w}{v})$ hexadacane and $1\%(\frac{w}{v})N{H}_3,$ which one is the limiting $(2)$ substrate?

$($b$)$ The reactor consists of $50L$ medium, assuming that $0\mathrm{.}3\frac{g}{L}$ hexadecane is the limiting substrate, i$.,$ what is the mass of bacterial cells $(g)$ generated?

ii$.$ what is the mass of $C{O}_2$ is produced from the complete conversion of hexadacane? Express your answer in units of $g.$

iii. how many cubic meter of $C{O}_2$ generated at $25C$ and $1$atm pressure?

$($c$)$ How many gram $($g$)$ of air is needed to sparge into the reactor to produce $15g$ of bacterial cells?

Answer: