$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (l)$ . If the total volume of the reactants is 27 L, calculate the volume of nitric oxide produced in the reaction.

Open in App

Solution

$4 N H_{3} (g) 4 vol. + 5 O_{2} (g) 5 vol. \to 4 N O (g) 4 vol. + 6 H_{2} O (l)$

Gay-Lussac’s law of combining volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure conditions remain constant. Let the volume of $N H_{3}$ be $4 x$ and that of $O_{2}$ be $5 x$ (because they react in the ratio of $4 : 5$ ). Volume of reactants given is 27 L; hence $9 x = 27 L$
So, $x = 3 L$
Volume of Nitric oxide produced = $4 x = 12 L$

Suggest Corrections

Similar questions

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

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

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

Δ_{f} H^{⊖}

N H_{3} (g)

N O (g)

H_{2} O (l)

- 11.04

+ 21.60

- 68.32 K c a l

4 N H_{3} (g) + 5 O_{2} (g) ⟶ 4 N O (g) + 6 H_{2} O (l)

{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)}

Join BYJU'S Learning Program