Mixtures of Ammonia and Oxygen

Question

A mixture of ammonia and oxygen is prepared by combining 300 mL of NH3 (measured at 759 torr and 28 Celsius) with 220 mL of O2 (measured at 780 torr and 50 Celsius). How many milliliters of N2 (measured at 740 torr an 100 Celsius) could be formed if the following reaction occurs?

4NH3 (g) + 3O2 (g) → 2N2 (g) + 6H2O (g)

Answer 1

4NH₃ (g) + 3O₂ (g) → 2N₂ (g) + 6H₂O (g)

moles NH₃ present = n = PV/RT = (759 torr)(0.3L)/(62.36 Ltorr/Kmol)(301K) = 0.01213 moles NH₃

moles O₂ present = n = PV/RT = (780 torr)(0.220L)/(62.36 Ltorr/Kmol)(323K) = 0.008519 moles O₂

Limiting reactant can be determined by dividing moles of each reactant by the coefficient in the balanced eq:

For NH₃: 0.01213 moles/4 = 0.0030

For O₂: 0.008519/3 = 0.00284 --> LIMITING REACTANT

Theoretical yield of N₂ = 0.008519 moles O₂ x 2 moles N₂/3 moles O₂ = 0.005679 moles N₂

Volume N₂ produced = nRT/P = (0.005670 mol)(62.36 Ltorr/Kmol)(373K)/740 torr) = 0.1785 L = 179 mls