The mass of ammonia in grams produced when 2.8 kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is _______.

Solution

Chemistry ma’am · 4 steps

Convert both masses to moles first, then let the limiting reagent decide how much ammonia forms.

1. N₂ + 3H₂ → 2NH₃

2. Moles of each. N₂: (2800 g)/(28 g mol⁻¹) = 100 mol. H₂: (1000 g)/(2 g mol⁻¹) = 500 mol.
3. Find limiting reagent. 100 mol of N₂ needs 3 × 100 = 300 mol of H₂. We have 500 mol, so H₂ is in excess and N₂ is limiting.
4. Ammonia formed. Each mole of N₂ gives 2 mol of NH₃, so 100 mol of N₂ gives 200 mol of NH₃. Its mass is 200 × 17 = 3400 g.
5. So. The mass of ammonia produced is 3400 g, and 200 mol of H₂ (400 g) is left unreacted.

The common slip: 3800. Adding the two masses, 2800 + 1000, gives 3800 g. That assumes all the hydrogen reacts, but 400 g of H₂ is left over.

The idea to keep. Whenever two reactant amounts are given, change both to moles and compare them using the equation's ratio. The limiting reagent, not the total mass, fixes the product.

Answer from the NTA final answer key. Solution by PrepWiser.