In the preparation of ${\mathrm{HNO}}_{3}$, we get $\mathrm{NO}$ gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of ${\mathrm{NH}}_{3}$ will be ......... .

1.  2

2.  3

3.  4

4.  6

Hint: $$\mathrm{NH}_{3} + \mathrm{O}_{2} \xrightarrow[Pt/Rh \ gauge \ catalyst]{\Delta } \mathrm{NO}(\mathrm{g})+ \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

When oxidation of NH3 takes place then NO is obtained as a product. The balanced equation is as follows:

$$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \xrightarrow[Pt/Rh \ gauge \ catalyst]{\Delta } 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$
From the reaction, it is clear that 4 moles of NH3 produce 4 moles of NO. Thus, 2 moles of NH3 produces 2 moles of NO.