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 $180L$ of water completely reacts with $N{O}_2$ produced to form nitric acid according to the above reactions, then the volume of air required at STP containing $20\%$ of $N{O}_2$ is:
(density of water is 1 g/mL)

• A. $1.56\times {10}^{6}L$
• B. $6.72\times {10}^{4}L$
• C. $3.36\times {10}^{6}L$
• D. $1.56\times {10}^{4}L$

Answer 1

The correct option is C $3.36\times {10}^{6}L$

Given , density of water is $1g/mL$
So, $180Lof{H}_{2}O=180\times {10}^{3}g{H}_{2}O$
$\therefore$ moles of water reacted = $\frac{180\times {10}^3}{18}={10}^{4}mol$

Again, from the sequence reaction,
1 mole of $H_{2}O$ reacts with 3 moles of $N{O}_2$
Thus, ${10}^{4}mol$ of water will react with $3\times {10}^4$ moles of $N{O}_2$.

$\therefore$, Volume of air required at STP
$=\frac{3\times {10}^4\times 22.4}{0.2}L=3.36\times {10}^{6}L$