Limiting reactants in moles

Question

What is the greatest amount of ammonia (NH3) in moles that can be made with 4.7 mol N2 and 6.9 mol H2? Which reactant is limiting ? How many moles of excess reactant will be left after the reaction ? 

Answer 1

Look at the balanced equation:

N₂ + 3H₂ ==> 2NH₃

To find the limiting reactant, we can determine how much NH3 is made with the given amounts of N₂ and H₂:

For N₂: 4.7 mol N₂ x 2 mol NH₃/mol N₂ = 9.4 moles NH₃ can be formed

For H₂: 6.9 mol H₂ x 2 mol NH₃/3 mol H₂ = 4.6 moles NH₃ can be formed. THUS H₂ IS LIMITING

Greatest amount of NH₃ that can be formed = 4.6 moles

Moles of excess reactant (N₂) is initial amount of N₂ - amount N₂ used in the reaction

initial amount N₂ = 4.7 moles

amount N₂ used in the reaction = 6.9 mol H₂ x 1 mol N₂/3 mol H₂ = 2.3 moles N₂ used

moles excess N₂ = 4.7 moles - 2.3 moles = 2.4 moles N₂ remains