INTRO TO LIMITING REACTANTS

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?

TO 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

PRACTICE WITH A CHEMICAL EQUATION

MnO2 + Al Mn + Al2O3

Balance the previous equation and then determine the limiting reactant if you have 2.5 grams of MnO2 and 5.4 grams of Al.

3MnO2 + 4Al 3Mn + 2Al2O3

2.5 g MnO2 x 1 mole MnO2 x 3 mol Mn = .028mol Mn 1 86.9 g MnO2 3 mol MnO2

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 MnO2. You can only make .028 moles of Mn if you react 2.5 g MnO2 and

5.4g Al .

ONE MORE PRACTICE PROBLEM

H2 + O2 H2O Balance the previous equation and determine

the limiting reactant if you start with 3g of H2 and 4 grams of O2

2H2 + O2 2H2O

3g of H2 x 1 mole H2 x 2 moles H2O = 1.5 mol H2O

1 2g H2 2 moles H2

4 g O2 x 1 mol O2 x 2 mol H2O = .25 mol H2O

1 32g O2 1 mol O2

The limiting reactant is oxygen.