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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.

Respuesta :

Answer:

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

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