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Limiting Reagent or Reactant

Nitric acid i...

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
$4 N H_{3} + 5 O_{2} (g) C a t a l y s t - ---- \to P t / R h 4 N O (g) + 6 H_{2} O (g) . . . (i)$
$2 N O (g) + O_{2} (g) 1120 K - --- \to 2 N O_{2} (g) . . . (i i)$
$3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (a q) + N O (g) . . . (i i i)$
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.

$85 k g$ of $N H_{3} (g)$ was heated with $320 k g$ oxygen in the first step and $H N O_{3}$ is prepared according to the above reactions. If the final solution has volume $500 L$ , then molarity of $H N O_{3}$ is:

2 M

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8 M

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3.33 M

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6.66 M

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Solution

The correct option is D 6.66 M
We know,
Molar mass of $N H_{3}$ = $17 g$
$∴ 85 k g N H_{3} = \frac{85000}{17} m o l = 5 \times 10^{3} m o l$
Molar mass of $O_{2}$ = $32 g$
$∴ 320 k g O_{2} = \frac{320000}{32} m o l = 10^{4} m o l$
Now, the given sequence of reactions are
$4 N H_{3} + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g)$
$2 N O (g) + O_{2} (g) 1120 K - --- \to 2 N O_{2} (g))$
$3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (a q) + N O (g)$

So, $4 m o l o f N H_{3}$ reacts with 5 mol of $O_{2}$
Hence $5 \times 10^{3}$ mol of $N H_{3}$ will react with $6.25 \times 10^{3} m o l$ 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 $N O$

Again 2 moles $N O$ gives 2 moles of $N O_{2}$
Thus $5 \times 10^{3}$ moles of $N O$ 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 $H N O_{3}$ , thus $5 \times 10^{3}$ moles of $N O_{2}$ will produce $\frac{2}{3} \times 5 \times 10^{3}$ moles of $H N O_{3}$

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

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Similar questions

4 N H_{3} + 5 O_{2} (g) C a t a l y s t - ---- \to P t / R h 4 N O (g) + 6 H_{2} O (g) . . . (i)

2 N O (g) + O_{2} (g) 1120 K - --- \to 2 N O_{2} (g) . . . (i i)

3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (a q) + N O (g) . . . (i i i)

68.5 %

98 %

85 k g

N H_{3} (g)

320 k g

H N O_{3}

500 L

H N O_{3}

4 N H_{3} + 5 O_{2} (g) C a t a l y s t - ---- \to P t / R h 4 N O (g) + 6 H_{2} O (g) . . . (i)

2 N O (g) + O_{2} (g) 1120 K - --- \to 2 N O_{2} (g) . . . (i i)

3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (a q) + N O (g) . . . (i i i)

68.5 %

98 %

85 k g

N H_{3} (g)

320 k g

H N O_{3}

500 L

H N O_{3}

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