Answer:
O2 is the limiting reactant.
Explanation:
Step 1: Data given
Mass of NH3 = 2.00 grams
Mass of O2 = 2.50 grams
Molar mass NH3 = 17.03 g/mol
Molar mass O2 = 32 g/mol
Step 2: The balanced equation
4NH3(g) + 5O2 (g) → 4NO(g) + 6H2O (g)
Step 3: calculate moles NH3
Moles NH3 = mass NH3 / molar mass NH3
Moles NH3 = 2.00 grams / 17.03 g/mol
Moles NH3 = 0.117 moles
Step 4: Calculate moles O2
Moles O2 = mass / molar mass O2
Moles O2 = 2.50 grams / 32 g/mol
Moles O2 = 0.0781 moles
Step 5: Calculate the limiting reactant
For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed. (0.0781 moles). NH3 is in excess. There will react 4/5 * 0.0781 moles = 0.0625 moles
There will remain 0.117 - 0.0625 = 0.0545 moles NH3
O2 is the limiting reactant.
