Ncert Exercises & Solutions

45.4 L of dinitrogen reacts with 22.7 L of dioxygen to produce 45.4 L of nitrous oxide as per the reaction:
$2 \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{~N}_2 \mathrm{O}(\mathrm{~g})$
Which of the following statements regarding this reaction are correct?

(a)	The reaction follows the Law of Gaseous Volumes.
(b)	The volume ratio of reactants (N₂: O₂) is 2:1.
(c)	The reaction violates the Law of Conservation of Mass.
(d)	The product N₂O has a volume equal to the volume of O₂ used.

Options:
1. (a) and (b)
2. (a) and (d)
3. (b) and (c)
4. (c) and (d)

Hint: The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction.

Explanation:

Step 1: The complete reaction is:

2 N_{2} (g) + O_{2} (g) \to 2 N_{2} O (g)

$2N_2(g) \ \ + \ \ O_2(g) \ \ \rightarrow\ \ 2N_2O(g)$
45.4 L 22.7L 45.4 L

So, the ratio of the volume of gases is:

V_{N_{2}} : V_{O_{2}} : V_{N_{2} O}

$V_{N_2} : V_{O_2} : V_{N_2O}$ =

2 : 1 : 2

$2:1:2$

Step 2: The gases are present in a simple whole number ratio. According to Gay Lussac's law, gases combine or are produced in a chemical reaction in a simple ratio by volume, provided that all gases are at the same temperature and pressure.

Hence, the given reaction follows Gay Lussac's law.

Step 3: This reaction does not violate the law of conservation of mass. As per the stoichiometry, one mole of

O_{2}

$O_2$ reacts with two moles of

N_{2}

$N_2$ to form two moles of

N_{2} O

$N_2O$ .

Thus, the volume of product formed is not equal to the volume of

O_{2}

$O_2$ consumed.

Therefore, option 1 is the correct answer.

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