Step 1
The reaction:
N2(g) + 3H2(g) → 2NH3(g) (balanced and completed)
All gases are assumed to be ideal and to be at STP conditions.
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Step 2
STP conditions:
1 mole of gas = 22.4 L
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Step 3
Information provided:
7 L of NH3 produced (ammonia)
Procedure:
By stoichiometry,
1 mole H2 = 22.4 L
1 mole NH3 = 22.4 L
N2(g) + 3H2(g) → 2NH3(g)
3 x 22.4 L H2 -------- 2 x 22.4 L NH3
X -------- 7 L NH3
X = 7 L NH3 x 3 x 22.4 L H2/2 x 22.4 L NH3 = 10.5 L
Answer: 10.5 L of H2 are consumed.