In the reaction, $4{\mathrm{NH}}_3\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4\mathrm{NO}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right),$ when 1 mole of ammonia and 1 mole of ${\mathrm{O}}_2$ are made to react to completion

$\begin{array}{l} 4\mathrm{N}{{\mathrm{H}}_3}_{\left(\mathrm{g}\right)}+5{{\mathrm{O}}_2}_{\left(\mathrm{g}\right)}\to 4{\mathrm{NO}}_{\left(\mathrm{g}\right)}+6{\mathrm{H}}_2\mathrm{O}{}_{\left(\mathrm{g}\right)}\\ 4\mathrm{moles} 5\mathrm{moles}\\ \mathrm{given} 1\mathrm{mole} ?\\ \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{O}}_2 \mathrm{required} = \frac{5}{4}>1\\ \mathrm{But} \mathrm{given} \mathrm{no}.\mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{Oxygen}= 1\\ \mathrm{So} {\mathrm{O}}_2 \mathrm{is} \mathrm{Limiting} \mathrm{reagent} \& \mathrm{it} \mathrm{consumes} \mathrm{completely}\end{array}$

From equation,

  $$\begin{gathered} 5\mathrm{moles} {\mathrm{ofO}}_2\to 4 \mathrm{moles} \mathrm{of} \mathrm{NO} + 6 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \\ 1 \mathrm{mole} \to ? ? \\ {\mathrm{n}}_{\mathrm{NO}} = \frac{4}{5}=0.8 ; {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}= \frac{6}{5}=1.2 \end{gathered}$$

  5 moles of ${\mathrm{O}}_2$ react with 4 moles of ${\mathrm{NH}}_3$

1 mole of  $O_2$   react with       ?

${\mathrm{n}}_{{\mathrm{NH}}_3}=\frac{4}{5}=0.8$

${\mathrm{n}}_{{\mathrm{NH}}_3}$left = 1-0.8 =0.2