1. Finding the Amount of Excess Reactant Left Over
2.00 g of NH3 reacts with 4.00 g of O2 according to the equation:
4 NH O2 → 4 NO + 6 H2O
2.00 g g
ER LR
If oxygen is the limiting reactant, how much excess reactant (NH3) will be left over at the end of the reaction?
STEPS:
Calculate the amount of excess reactant that will be used up
Set up a stoichiometry problem
Start with the LIMITING REACTANT (LR) and solve for the EXCESS REACTANT (ER) that will be used
2. Subtract the amount of excess reactant used from the amount of excess reactant given to find the amount of excess reactant left over.

2. 4 NH O2 → 4 NO + 6 H2O
2.00 g g
ER LR
Find amount of NH3 used
4.00 g O2
1 mol O2
4 mol NH3
17.03 g NH3
= 1.70 g NH3 USED
32.00 g O2
5 mol O2
1 mol NH3
2) Subtract the amount of NH3 used from the amount of NH3 given in the problem to find the amount of NH3 left over
2.00 g – 1.70 g = 0.30 g NH3 left over

3. 4 Na (s) + O2(g)  2 Na2O (s)
5.00 g of sodium reacted with 5.00 g of oxygen
a. How many grams of product can form?
b. What is the limiting reactant?
c. How much excess reactant is left over? (hint: first find the amount of excess reactant used in the reaction, and subtract from the amount given)

4. 4 Na + O2 → 2 Na2O
5.00 g Na
1 mol Na
2 mol Na2O
61.98 g Na2O
= 6.74 g Na2O
22.99 g Na
4 mol Na
1 mol Na2O
5.00 g O2
1 mol O2
2 mol Na2O
61.98 g Na2O
= 19.4 g Na2O
32.00 g O2
1 mol O2
1 mol Na2O
Theoretical Yield = 6.74 g of Na2O; Na is the Limiting Reactant; O2 is the excess reactant
32.00 g O2
5.00 g Na
1 mol Na
1 mol O2
= 1.74 g of O2 USED
22.99 g Na
4 mol Na
1 mol O2
5.00 g O2 (given) g O2 (used) = g O2 LEFT OVER