Question

3. 56 g of N₂ reacts 9 g of H₂ in a closed vessel to form NH3. Which reactant is left in excess and how 2207 much? 1) N₂, 10. g +3th = 2013/ (3) N₂, 14 g (2) H₂, 4 g (4) H₂,3 g N₂- to d A​

Answer 1

Answer: the reactant left in excess is H₂, and the amount of excess H₂ is 4.5 moles - 3 moles = 1.5 moles.

Explanation: Moles of N₂ = mass of N₂ / molar mass of N₂

Moles of N₂ = 56 g / 28 g/mol = 2 moles

Moles of H₂ = mass of H₂ / molar mass of H₂

Moles of H₂ = 9 g / 2 g/mol = 4.5 moles