Question

N2 (g) + 3 H2 (g) → 2 NH3 (g) A 56 kg sample of N2 (g) and a 9.0 kg sample of H2 (g) are combined in a closed vessel, where they react according to the equation above. After the reaction has proceeded to completion, how many moles of NH3 (g) have been produced?

Answer 1

Using stoichiometry, 56 kg of N₂ and 9 kg of H₂ will produce 4,000 moles of NH₃ when the reaction N₂(g) + 3H₂(g) → 2NH₃(g) goes to completion, with nitrogen being the limiting reactant.

Step-by-step explanation:

The stoichiometry of the chemical reaction between nitrogen and hydrogen gases to produce ammonia can be used to determine the amount of ammonia produced. According to the balanced chemical equation, N₂(g) + 3H₂(g) → 2NH₃(g), 1 mole of N₂ reacts with 3 moles of H₂ to give 2 moles of NH₃.

To find the number of moles of ammonia produced, we first need to convert the masses of N₂ and H₂ to moles using their molar masses (N₂ = 28 g/mol, H₂ = 2 g/mol).

For 56 kg N₂, we have (56,000 g) / (28 g/mol) = 2,000 moles of N₂. For 9.0 kg H₂, we have (9,000 g) / (2 g/mol) = 4,500 moles of H₂. Since the reaction requires 3 moles of H₂ for every 1 mole of N₂, hydrogen is in excess, and nitrogen will be the limiting reactant.

Thus, 2,000 moles of N₂ will produce (2,000 moles N₂) x (2 moles NH₃) / (1 mole N₂) = 4,000 moles of NH₃.