Question

In Ostwald process of manufacturing of $HN{O_3}$ Catalyst used is:
(A).$MO$
(B). $Fe$
(C). $Mn$
(D). $Pt$

Answer 1

Hint:Ostwald forces is the preaches of conversion of ammonia $\left( {N{H_3}} \right)$ into nitric acid $\left( {HN{O_3}} \right)$ by firstly oxidation and then reacting with water.

Complete step by step answer:
We know that the Ostwald process in the process of manufacture of nitric acid $\left( {HN{O_3}} \right)$ this process was discovered by Wilhelm Ostwald in 1902. The Ostwald process is carried out in the chemical industry ro produce the main few materials for the most common type of fertilizer production. The Ostwald process is carried out as follow :
First of all ammonia is taken and is oxidized in the catalytic chamber at ${600^0}$C temperature and in presence of platinum catalyst. This leads to the formation of nitric oxide. The process is reversible and is endothermic.
$4N{H_3} + 5{O_2}\xrightarrow{{PT}}4NO + 6{H_2}O$
The nitric oxide (NO) produced is again passed through a heat exchanger to lower its temperature. After cooling, nitric acid is passed through another oxidizing tower where nitrogen dioxide is formed at about ${50^0}$
$2NO + {O_2} \to 2N{O_2}$
Now the released nitrogen dioxide is treated with water which produces our required nitric acid $\left( {H{NO_3}} \right)$ as product
$3N{O_2} + {H_2} \to 2HN{O_3} + NO$
Since in the whole reaction the only catalyst used is platinum hence Pt catalyst is used in the Ostwald process.
Hence option (D) is the correct answer.

Note:
The reaction process is exothermic as energy is released. So sometimes producing $HN{O_3}$ by the Ostwald process can be dangerous due to some unfavorable condition, The most insignificant condition is the corrosive nature of nitric acid.