First slide

Gaseous state

Question

If 1 L of NH₃ is formed at STP according to reaction,
$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3}$
moles of H₂ required under identical condition would be

Moderate

A

3 mol
B

1.5 mol
C

0.067 mol
D

0.134 mol

Solution

$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3} 1 mol 3 mol 2 mol 1 L 3 L 2 L$
$2 L {NH}_{3} is from = 3 {LH}_{2}$
$∴ 1 L {NH}_{3} is from = 1 .5 {LH}_{2}$
$= \frac{1 .5 L}{22 .4 L {mol}^{- 1}} = 0 .067 mol$

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