Nitric acid can be produced from ammonia in a three steps process-4NH3 g-5O2 g- 4NOg-6H2 Og i2NOg-O2 g- 2NO2 g ii3NO2 g-H2 O - 2HNO3 aq-NOg iiiGiven the percentage yields of i- ii and iii reactions as 50

4NH_3 (g)+5O_2 (g)→ 4NO(g)+6H_2 O(g) (i) 2NO(g)+O_2 (g)→ 2NO_2 (g) (ii) 3NO_2 (g)+H_2 O → 2HNO_3 (aq)+NO(g) (iii) Moles of HNO_3 produced = 157563= 25 mol In ...

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Standard XII

Chemistry

Stoichiometric Calculations

Nitric acid c...

Question

Nitric acid can be produced from ammonia in a three steps process:
$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g) (i)$
$2 N O (g) + O_{2} (g) \to 2 N O_{2} (g) (ii)$
$3 N O_{2} (g) + H_{2} O \to 2 H N O_{3} (a q) + N O (g) (iii)$
Given the percentage yields of (i), (ii) and (iii) reactions as 50 %, 60 % and 80 %. What will be the volume of $N H_{3}$ (g) at STP required to produce 1575 g of $H N O_{3}$ .

156.25 L

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350 L

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3500 L

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Solution

The correct option is C 3500 L
$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g) (i)$
$2 N O (g) + O_{2} (g) \to 2 N O_{2} (g) (ii)$
$3 N O_{2} (g) + H_{2} O \to 2 H N O_{3} (a q) + N O (g) (iii)$
Moles of $H N O_{3}$ produced = $\frac{1575}{63}$ = 25 mol
In reaction (iii),
2 mol of $H N O_{3}$ is produced by 3 mol of $N O_{2}$
25 mol of $H N O_{3}$ will be produced by $= \frac{3}{2} \times 25 = 37.5$ mol of $N O_{2}$
Yield of the reaction is given as 80 %
Actual amount of $N O_{2}$ required = $\frac{37.5}{80} \times 100$ = 46.875 mol
In reaction (ii),
2 mol of $N O_{2}$ is produced from 2 mol of $N O$
46.875 mol of $N O_{2}$ will be formed by 46.875 mol of $N O$
Yield of the reaction is given as 60 %
Actual amount of NO required = $\frac{46.875}{60} \times 100$ = 78.125 mol
In reaction (i),
4 mol of $N O$ are produced by 4 mol of $N H_{3}$
78.125 mol will be produced by 78.125 mol of $N H_{3}$
Yield of the reaction is given as 50 %
Actual amount of $N H_{3}$ required = $\frac{78.125}{50} \times 100$ = 156.25 mol
1 mol of $N H_{3}$ occupies a volume of 22.4 L at STP
156.25 mol of $N H_{3}$ will occupy = 3500 L

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Similar questions

Q. Nitric acid can be produced from ammonia in a three steps process:

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

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

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

Given the percentage yields of (i), (ii) and (iii) reactions as 50 %, 60 % and 80 %. What will be the volume of

N H_{3}

(g) at STP required to produce 1575 g of

H N O_{3}

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}

If 1 mole of aqueous nitric acid is formed, calculate total heat released :

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

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

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

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.

If

180 L

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 %

N O_{2}

is:
(density of water is 1 g/mL)

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

180 L

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 %

N O_{2}

is:
(density of water is 1 g/mL)

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