1. I NTRO TO L IMITING REACTANTS

2. a) How many cars can be assembled from the following parts: 140 car bodies, 520 tires, 270 headlights. b) What is left over? c) What was the “limiting part”? d) What part(s) were in excess?

3. T O DETERMINE LIMITING REACTANT Use stoichiometry to determine how much of one product you can get from each of the reactants (ex. How many cars you can make from 520 tires, how many cars you can make from 140 car bodies, etc) Whichever reactant gives you the least amount of product is your limiting reactant

4. P RACTICE WITH A CHEMICAL EQUATION MnO 2 + Al  Mn + Al 2 O 3 Balance the previous equation and then determine the limiting reactant if you have 2.5 grams of MnO 2 and 5.4 grams of Al.

5. 3MnO 2 + 4Al  3Mn + 2Al 2 O 3 2.5 g MnO 2 x 1 mole MnO 2 x 3 mol Mn =.028mol Mn 1 86.9 g MnO 2 3 mol MnO 2 5.4g Al x 1 mole Al x 3 mol Mn =.150 mol Mn 1 26.9g Al 4 mol Al The limiting reactant is MnO 2. You can only make.028 moles of Mn if you react 2.5 g MnO 2 and 5.4g Al.

6. O NE MORE PRACTICE PROBLEM H 2 + O 2  H 2 O Balance the previous equation and determine the limiting reactant if you start with 3g of H 2 and 4 grams of O 2

7. 2H 2 + O 2  2H 2 O 3g of H 2 x 1 mole H 2 x 2 moles H 2 O = 1.5 mol H 2 O 1 2g H 2 2 moles H 2 4 g O 2 x 1 mol O 2 x 2 mol H 2 O =.25 mol H 2 O 1 32g O 2 1 mol O 2 The limiting reactant is oxygen.