First slide

Gaseous state

Question

A mixture of 50.0 mL of NH₃ and 60.0 mL of O₂ gas reacts as
$4 {NH}_{3} (g) + 5 O_{2} (g) ⟶ 4 NO + 6 H_{2} O (g)$
If all the gases are at the same temperature and pressure and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced?

Moderate

A

48 mL
B

60.0 mL
C

72 mL
D

75.0 mL

Solution

By Avogadro's law, equal volumes of gases contain equal number of moles.
$4 {NH}_{3} (g) + 5 O_{2} (g) ⟶ 4 NO (g) + 6 H_{2} O (g) 4 mL {NH}_{3} = 6 {mLH}_{2} O$
$∴ 50 mL {NH}_{3} = 75 {mLH}_{2} O 5 {mLO}_{2} = 6 {mLH}_{2} O ∴ 60 {mLO}_{2} = 72 {mLH}_{2} O stoichiometrically, O_{2} is consumed completely Thus, 72 {mLH}_{2} O (g) .$

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