If 20 ml of NH 3 is mixed with 100 ml of O 2 . Volume contraction at the completion of above reactions is

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If 20 ml of NH 3 ​ is mixed with 100 ml of O 2 ​ . Volume contraction at the completion of above reactions is

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Question Explanation: Calculating volume contraction in ammonia oxidation.

Concept: Stoichiometry and Limiting Reactant. Solution: The question implies the complete oxidation of ammonia to $NO_{2}$ and liquid water. First, write the overall balanced chemical equation. Then, identify the limiting reactant and calculate the initial and final total volumes of gases to find the contraction.

Calculation:

(1) $4 N H_{3} (g) + 5 O_{2} (g) ⟶ 4 NO (g) + 6 H_{2} O (l)$

$2 NO (g) + O_{2} (g) ⟶ 2 N O_{2} (g)$

(2) $4 NO (g) + 2 O_{2} (g) ⟶ 4 N O_{2} (g)$

Adding (1) and (2) gives the overall reaction:

$4 N H_{3} (g) + 7 O_{2} (g) ⟶ 4 N O_{2} (g) + 6 H_{2} O (l)$

Initial volumes:

$N H_{3}$ : 20 ml

$O_{2}$ : 100 ml

Since 4 volumes of $N H_{3}$ react with 7 volumes of $O_{2}$ , 20 ml of $N H_{3}$ will react with $20 \times \frac{7}{4} = 35$ ml of $O_{2}$ .

$N H_{3}$ is the limiting reactant.

After reaction:

$N H_{3}$ : 0 ml

$O_{2}$ : $100 - 35 = 65$ ml (remaining)

$N O_{2}$ : 20 ml (since 4 volumes of $N H_{3}$ produce 4 volumes of $N O_{2}$ )

$H_{2} O (l)$ : liquid, so its volume is negligible in terms of gas volume.

Initial total volume of gases ( $V_{i}$ ):

$V_{i} = 20 m l (N H_{3}) + 100 m l (O_{2}) = 120 m l$

Final total volume of gases ( $V_{f}$ ):

$V_{f} = 65 m l (O_{2}) + 20 m l (N O_{2}) = 85 m l$

Volume contraction ( $V_{co n t r a c t i o n}$ ):

$V_{co n t r a c t i o n} = V_{i} - V_{f} = 120 m l - 85 m l = 35 m l$

Final Answer: 35 ml

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