Question

How many grams of nitrogen will be produced in a reaction between ammonia and dinitrogen monoxide in which the total heat energy released was 505 KJ?

2 NH3 + 3 N2O = 4 N2 + 3 H2O DELTA Hrxn= ?
Given:
4 NH3 + 3 O2 = 2 N2 + 6 H2O DELTA Hrxn= -1556KJ
N2O + H2 = N2 + H2O DELTA Hrxn= -389.4 KJ
H2+ ½ O2 = H2O DELTA Hrxn= -223.9 KJ

Answer 1

55.5g of N₂

Step-by-step explanation:

It is possible to find ΔH of a reaction by the sum of half-reaction (Hess's law), thus:

(1) 4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O ΔH= -1556KJ

(2) N₂O + H₂ → N₂ + H₂O ΔH= -389.4 KJ

(3) H₂+ ½ O₂ → H₂O ΔH= -223.9 KJ

(4) = 1/2 (1) + 3 (2):

(4) 2 NH₃ + 3 N₂O +3 H₂ + ³/₂ O₂ → 4 N₂ + 6 H₂O

ΔH = 1/2 (-1556kJ) + 3 (-389.4kJ) = -1946.2kJ

(4) - 3(3):

2 NH₃ + 3 N₂O → 4 N₂ + 3 H₂O ΔH = -1946.2kJ - 3 (-223.9 KJ)

ΔH = -1274.5 kJ

As when 5 moles of nitrogen are produced there are released 1274.5kJ, when are released 505kJ:

505kJ × (5 moles N₂ / 1274.5kJ) = 1.98 moles of N₂. In mass:

1.98moles N₂ ₓ (28g / 1mol) = 55.5g of N₂