1. Limiting Reactant

2. What is the significance of the limiting reactant?
Limiting reactant (LR) tells you the max amount of product that can be produced
LR is always completely used up
** What happened in the smores lab?? What was our limiting reactant?!?!

3. What happens to the amount of reactant that is present in excess?
Called the excess reactant (ER)
The ER will not react and be left over

4. Steps to Calculating Product from the LR
Find the number of moles of both reactants
Determine the number of moles of one product made from each reactant
Identify the LR by the reactant that produces the smaller amount of moles for the product
Use the LR to find the mass of the product

5. Example #1 Ethane undergoes the following combustion reaction:
C2H O H2O CO2
If 2.70 moles C2H4 is reacted with 6.30 mole of O2, identify the limiting reactant and determine the number of moles of carbon dioxide produced.

6. Example #1 (continue) C2H4 + 3 O2 2 H2O + 2 CO2 2.70 mole C2H4
6.30 mole O2 x 2 moles CO2 = x
2 moles CO2
5.4 moles CO2 =
1 mole C2H4
LR = O2
4.2 moles CO2
3 mole O2

7. Example #2
Hydrogen gas can be produced by the reaction of magnesium metal with hydrochloric acid. Identify the limiting reactant when 6.00 g of HCl reacts with 5.00 g of Mg and determine the mass of hydrogen produced.
Mg HCl MgCl H2

8. Example #2 Mg + 2 HCl MgCl2 + H2 6.00 g HCl x 5.00 g Mg x
moles H2 x
1 mole HCl x
1 mole H2 =
mole H2
36.46 g HCl
2 mole HCl
LR = HCl
1 mole Mg
1 mole H2 x =
0.206 mole H2
24.31 g Mg
1 mole Mg
2.02 g H2 =
0.166 g H2
1 mole H2

9. Example #3
Phosphoric acid is produced by the following reaction. What is the maximum number of grams of H3PO4 that can be formed when g of P4O10 reacts with g H2O?
P4O H2O H3PO4

10. Example #3 P4O10 + 6 H2O 4 H3PO4 100 g P4O10 x 200 g H2O x
1.41 moles H3PO4 x
1 mole P4O10 x
4 mole H3PO4 =
1.41 moles H3PO4
283.9 g P4010
1 mole P4O10
LR = P4O10
1 mole H2O
4 mole H3PO4 x =
7.40 moles H3PO4
18.02 g H2O
6 mole H2O
98.00 g H3PO4 =
138.1 g H3PO4
1 mole H3PO4