If 3.52 L of nitrogen gas and 2.75 L of hydrogen gas were allowed to react, how many litres of ammonia gas could form? Assume all gases are at the same temperature and pressure.

V NH3 (g) = 1.833 L

Explanation:

balanced reaction:

N2 (g) + 3H2 (g) → 2NH3 (g)

assuming STP:

∴ V N2 (g) = 3.52 L

∴ V H2 (g) = 2.75 L

ideal gas:

PV = RTn

∴ moles N2 (g) = PV/RT

⇒ mol N2 (g) = (1 atm) (3.52 L) / (0.082 atm. L/K. mol) (298 K)

⇒ mol N2 (g) = 0.144 mol

∴ moles H2 (g) = PV/RT

⇒ mol H2 (g) = (1) (2.75) / (0.082) (298) = 0.113 mol (limit reagent)

∴ moles NH3 (g) = (0.113 moles H2 (g)) (2 moles NH3 / 3 mol H2) = 0.075 mol

∴ V NH3 (g) = RTn/P

⇒ V NH3 (g) = ((0.082 atm. L/K. mol) (298 K) (0.075 mol)) / (1 atm)

⇒ V NH3 (g) = 1.833 L