According to Gay-Lussac's Law of Combining Volumes, what volume of ammonia gas (NH_3) is p… — JAMB UTME

According to Gay-Lussac's Law of Combining Volumes, what volume of ammonia gas ( $NH_3$ ) is produced when $20\text{ cm}^3$ of nitrogen gas ( $N_2$ ) reacts completely with excess hydrogen gas ( $H_2$ ) at constant temperature and pressure?

A
$10\text{ cm}^3$
B
$20\text{ cm}^3$
$40\text{ cm}^3$ Answer
D
$60\text{ cm}^3$

Answer

40\text{ cm}^3

of ammonia gas (

NH_3

) is produced.

According to the balanced chemical equation

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

, 1 mole (or volume) of

N_2

reacts to form 2 moles (or volumes) of

NH_3

. Therefore,

20\text{ cm}^3

N_2

produces

2 \times 20\text{ cm}^3 = 40\text{ cm}^3

NH_3

gas.

Step-by-Step Solution

Write the balanced chemical equation for the synthesis of ammonia

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

Stoichiometric coefficients determine the combining volume ratios of reacting and product gases under constant temperature and pressure.

Determine the volume ratio between

N_2

and

NH_3

1\text{ volume of } N_2 : 2\text{ volumes of } NH_3

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number ratios by volume.

Calculate the volume of

NH_3

formed from

20\text{ cm}^3

N_2

\text{Volume of } NH_3 = 20\text{ cm}^3 \times 2 = 40\text{ cm}^3

Directly scaling the volume of

N_2

by the coefficient ratio (2/1).

Key Concept

Gay-Lussac's Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volume of the product if gaseous, provided temperature and pressure remain constant.

Estimated Time:45s

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