Question

In the preparation of $HN{O_3}$, we get $NO$ gas by catalytic oxidation of ammonia. The moles of $NO$ produced by the oxidation of two moles of $N{H_3}$ will be.
A) $2$
B) $3$
C) $4$
D) $6$

Answer 1

We know that the Ostwald process is one among the foremost common methods used for the manufacturing of sulfuric acid and this method was developed within the year 1902 by a German chemist named Ostwald.

Complete step by step answer:
In the manufacturing process of nitric acid, we get No gas by catalytic oxidation of ammonia. In the below equation four moles of ammonia produced four moles of NO. So the moles of NO produced by the oxidation of two moles of ammonia will be two moles.
We can write the chemical equation for the reaction takes place at this step as,
$4N{H_3} + 5{O_2} \to 4NO + 6{H_2}O$

So, the correct answer is Option A.

Note:
Generally, within the Ostwald process ammonia is transformed into liquid ammonia. Vanadium pentoxide $\left( {{V_2}{O_5}} \right)$ is usually used as a catalyst during the processing of sulfuric acid during this method. There are several steps involved. We shall discuss them intimately below.

Step 1: Catalytic oxidation of Ammonia:
A mixture of parched air and parched ammonia within the ratio of $10:1$ by volume is compressed then passed into platinum gauze which acts as catalyst at about.
The chemical equation for this step is given as below,
$4N{H_3}{\text{ }} + 5{O_2}\xrightarrow{{}}4NO + 6{H_2}O + Heat$

Step 2: Oxidation of gas:
Nitric oxide combines with oxygen to make dioxide at about $50^\circ C$.
The reaction takes place at this stage is write as,
$2NO + {O_2}\xrightarrow{{}}2N{O_2}$
Step 3: Absorption of dioxide in water:
The dioxide and oxygen present within the air react with water to make sulfuric acid.
The reaction takes place at this step is given as,
$4N{O_2} + 2{H_2}O + {O_2}\xrightarrow{{}}4HN{O_3}$
Nitric acid obtained is concentrated above $50\%$. On further distillation, $68\%$ of sulfuric acid is produced.