Question

Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major industrial chemicals and is widely used. Nitric acid is manufactured by the catalytic oxidation of ammonia that is known as Ostwald Process which can be represented by sequence of reactions shown below
$4N{H}_3+5O_2\left(g\right)\underset{Pt/Rh}{\overset{Catalyst}{\to }}4NO\left(g\right)+6H_{2}O\left(g\right)...\left(i\right)$
$2NO\left(g\right)+O_2\left(g\right)\stackrel{1120K}{\to }2N{O}_2\left(g\right)...\left(ii\right)$
$3N{O}_2\left(g\right)+H_{2}O\left(l\right)\to 2HN{O}_3\left(aq\right)+NO\left(g\right)...\left(iii\right)$
The aqueous nitric acid obtained by this method can be concenteated by distillation to $68.5\%$ by weight. Further concentration to $98\%$ acid can be achieved by dehydration with concentrated sulphuric acid.

If $170kg$ of $N{H}_3$ is heated in excess of oxygen, then the volume of $H_{2}O\left(l\right)$ produced in $1^{st}$ reaction at STP is (Given that density of water is $1\frac{g}{ml}$)

• A. $336\times {10}^{3}L$
• B. $270L$
• C. $224\times {10}^{3}L$
• D. $170L$

Answer 1

The correct option is B $270L$

Moles of $N{H}_3$ = $\frac{170000}{17}=10\times {10}^{3}mol$
Moles of $H_{2}O$ formed = $\frac{6}{4}\times 10\times {10}^{3}mol$
$\therefore$ Mass of $H_{2}O$ formed $=\frac{6}{4}\times 10\times {10}^3\times 18=270\times {10}^{3}g$
As the density of water is 1 g per ml, then volume of $H_{2}O$ formed $=270\times {10}^{3}ml=270L$