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.

$85kg$ of $N{H}_3\left(g\right)$ was heated with $320kg$ oxygen in the first step and $HN{O}_3$ is prepared according to the above reactions. If the final solution has volume $500L$, then molarity of $HN{O}_3$ is:

• A. 2 M
• B. 8 M
• C. 3.33 M
• D. 6.66 M

Answer 1

The correct option is D 6.66 M

We know,
Molar mass of $N{H}_3$ = $17g$
$\therefore 85kgN{H}_3=\frac{85000}{17}mol=5\times {10}^{3}mol$
Molar mass of $O_2$ = $32g$
$\therefore 320kg{O}_2=\frac{320000}{32}mol={10}^{4}mol$
Now, the given sequence of reactions are
$4N{H}_3+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$
$2NO\left(g\right)+O_2\left(g\right)\stackrel{1120K}{\to }2N{O}_2\left(g\right))$
$3N{O}_2\left(g\right)+H_{2}O\left(l\right)\to 2HN{O}_3\left(aq\right)+NO\left(g\right)$

So, $4molofN{H}_3$ reacts with 5 mol of $O_2$
Hence $5\times {10}^3$ mol of $N{H}_3$ will react with $6.25\times {10}^{3}mol$ of $O_2$
So, $N{H}_3$ is limiting reagent here.

From the above sequence of reaction, $5\times {10}^3$ moles of $N{H}_3$ will produce $5\times {10}^3$ moles of $NO$

Again 2 moles $NO$ gives 2 moles of $N{O}_2$
Thus $5\times {10}^3$ moles of $NO$ will give $5\times {10}^3$ moles of $N{O}_2$.

From the reaction sequence, 3 moles of $N{O}_2$ produces 2 moles of $HN{O}_3$, thus $5\times {10}^3$ moles of $N{O}_2$ will produce $\frac{2}{3}\times 5\times {10}^3$ moles of $HN{O}_3$

Given, final solution has volume = $500L$
So, molarity = $\frac{\frac{2}{3}\times 5\times {10}^3}{500}M=6.66M$