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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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None of these

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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

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)

N H_{3}

H N O_{3}

N H_{3}

(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)

%

i, i i

i i i

50 %, 60 %

80 %

N H_{3} (g)

1 a t m

0^{\circ} C

1575 g

H N O_{3}

{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)}

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)

(i n kJ/mole)

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 %

180 L

N O_{2}

20 %

N O_{2}

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