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If 4.71 l of nitrogen gas and 24.96 l of hydrogen gas were allowed to react, how many liters of ammonia gas could form? assume all gases are at the same temperature and pressure.

Respuesta :

In total 16.616 L of ammonia could form.

Calculation

The balanced reaction for this would be:

N2(g) + 3H2(g) → 2NH3(g)

assuming STP:

∴ V N2(g) = 4.71 L

∴ V H2(g) = 24.96 L

The ideal gas equation is stated as:

PV = RTn

∴ moles N2(g) = PV/RT

⇒ mol N2(g) = (1 atm) (4.71L)/ (0.082 atm.L/K.mol)(298 K)

⇒ mol N2(g) = 0.007 mol

∴ moles H2(g) = PV/RT

⇒ mol H2(g) = (1) (24.96) / (0.082) (298) = 1.021 mol (limit reagent)

∴ moles NH3(g) = (1.021 moles H2(g))(2 moles NH3 / 3 mol H2) = 0.680 mol

∴ V NH3(g) = RTn/P

⇒ V NH3(g) = ((0.082 atm.L/K.mol)(298 K)(0.680 mol))/(1 atm)

⇒ V NH3(g) = 16.616 L

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