$=9\mathrm{.}43{10}^{-4}$ moles

$O_2$ nccded $=\frac{1}{2}329\mathrm{.}43{10}^{-4}=\mathrm{.}169=166Mg$

Problem

Example: What is the NBOD, in $m\frac{g_2}{l},$ of a waste containing $20$

$m\frac{g}{L}$ ammonia as nitrogen $(N{H}_3-N)?$

$($MW N $=14\frac{g}{m}ol)$

$232$ PC$$ mule

$N{H}_4^{+}{?}_{\{\begin{array}{ccc}\text{:}[& y& ]\\ 14\text{g}\text{m}\text{o}\text{l}\end{array}}+2O_{2}N{O}_3^{-}+H_{2}O+2H^{+},\frac{20\frac{ms}{L}}{14\frac{g}{L}}=x*(2\frac{}{2})=91ms$

My Answer is closest to:

A$.5mgg\frac{O_2}{l}$

B$.45mg\frac{O_2}{\text{I}}$

C$.91mg{O}_{2}1$