If 5.84 L of nitrogen gas and 24.96 L of hydrogen gas were allowed to react, how many liters of ammonia gas could form? ?

At constant temperature and pressure, volume ratio of gases is equal to their mole ratio.

N₂(g) + 3H₂(g) → 2NH₃(aq)
Volume ratio N₂ : H₂ = 1 : 3

If N₂ completely reacts, H₂ needed = (5.84 L) × 3 = 17.52 L < 24.96 L
Hence, H₂ is in excess, and the limiting reactant is N₂.

According to the above equation, volume ratio N₂ : NH₃ = 1 : 2
Volume of NH₃ formed = (5.84 L) × 2 = 11.68 L

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OR:
If H₂ completely reacts:
(5.84 L N₂) × (2 L NH₃ / 1 L N₂) = 11.68 L NH₃

If N₂ completely reacts:
(24.96 L H₂) × (2 L NH₃ / 3 L H₂) = 16.64 L NH₃

When completely reacts, the limiting reactant produces the smallest amount of product.
Hence, H₂ is the limiting reactant.
Volume of NH₃ formed = 11.68 L

coefficients of a balanced equation can be read as volume ratios for ideal gases at the same P and T.

balanced equation
.. 1 N2 + 3 H2 ---> 2 NH3

if N2 is the LR
.. 5.84L N2 * (2L NH3 / 1L N2) = 11.7L NH3

if H2 is the LR
.. 24.96L H2 * (2L NH3 / 3L H2) = 16.6L NH3

since N2 gave less NH3, N2 is the LR and the theoretical yield of NH3 = 11.7L