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 4NH3-5O2g4NOg-6H2Og -i 2NOg-O2g2NO2g -ii 3NO2g-H2Ol- 2HNO3aq-NOg -iii 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 170 kg of NH3 is heated in excess of oxygen- then the volume of H2Ol produced in 1st reaction at STP is Given that density of water is 1gml

The correct option is B 270 LMoles of NH_3 = 17000017=10× 10^3 mol nbsp; Moles of H_2O formed = 64× 10× 10^3 mol ∴ Mass of H_2O formed =64× 10× 10^3× 18 = 27 ...

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Standard XII

Chemistry

Stoichiometric Calculations

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.

If $170 k g$ of $N H_{3}$ is heated in excess of oxygen, then the volume of $H_{2} O (l)$ produced in $1^{s t}$ reaction at STP is (Given that density of water is $1 \frac{g}{m l}$ )

336 \times 10^{3} L

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270 L

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224 \times 10^{3} L

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170 L

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Solution

The correct option is B $270 L$
Moles of $N H_{3}$ = $\frac{170000}{17} = 10 \times 10^{3} m o l$
Moles of $H_{2} O$ formed = $\frac{6}{4} \times 10 \times 10^{3} m o l$
$∴$ 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 $= 270 \times 10^{3} m l = 270 L$

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Q. 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

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:

Q. Name the following:
(i) Gas produced in air during lightning.
(ii) Gas obtained by treating manganese with 1% nitric acid.
(iii) Salt referred to as Chile saltpetre.
(iv) Salt used in the laboratory to prepare nitric acid.
(v) Products obtained by the catalytic oxidation of ammonia.
(vi) Products obtained by heating concentrated nitric acid.
(vii) Two coloured nitrates.

Q. (a) Nitric acid cannot be concentrated beyond 68% by the distillation of a dilute solution of HNO₃. State the reason.
(b) How can 68% nitric acid be concentrated to 98% nitric acid?

Q. The following sequence of reaction occurs in commercial production of aqueous nitric acid.

4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (l) Δ H = - 904 k J . . . (1)

2 N O (g) + O_{2} (g) \to 2 N O_{2} (g) Δ H = - 112 k J . . . (2)

3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (a q) + N O (g) Δ H = - 140 k J . . . (3)

Determine the magnitude of total heat

(i n kJ/mole)

liberated at constant pressure for the production of exactly 1 mole of aqueous nitric acid by this process

Q. Nitric acid can be produced by

N H_{3}

in three steps process:

(i) 4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g)

(i i) 2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)

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

%

yield of

i, i i

and

i i i

are

50 %, 60 %

and

80 %

respectively, then what volume of

N H_{3} (g)

1 a t m

and

0^{\circ} C

required to produce

1575 g

H N O_{3}

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The correct option is B 270 LMoles of NH3 = 17000017=10×103 mol Moles of H2O formed = 64×10×103 mol ∴ Mass of H2O formed =64×10×103×18= 270×103 g As the density of water is 1 g per ml, then volume of H2O formed = 270×103 ml=270 L