1. Chapter 9 - Section 3 Suggested Reading: Pages 312 - 318
Stoichiometry
Chapter 9 - Section 3 Suggested Reading: Pages

2. Limiting Reactants Available Ingredients 4 slices of bread
1 jar of peanut butter
1/2 jar of jelly
Limiting Reactant
bread
Excess Reactants
peanut butter and jelly

3. Limiting Reactants Available Ingredients Copper Wire 0.5 g AgNO3
0.5 grams AgNO3
Excess Reactants
Copper Wire

4. Limiting Reactant
The reactant that limits the amount of product that can be formed.

5. When quantities of reactants are available in the exact ratio described by the balanced equation, they are said to be in stoichiometric proportions.

6. Limiting Reactants Limiting Reactant used up in a reaction
determines the amount of all products formed
Excess Reactant
added to ensure that the other reactant is completely used up
usually cheaper & easier to recycle

7. Solving Problems – Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the amount of product formed.
3. The reactant that produces the smaller amount of product is the limiting reactant.
Very similar to mass-mass problems!

8. Step 1: Write a balanced equation.
Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water.
O H2  H2O

9. Step 2:
For each reactant, calculate the amount of product formed.
Identify the limiting reactant when 1.22 g of oxygen reacts with 1.05 g of hydrogen to produce water.
O H2  H2O

10. = O2 + 2H2  2 H2O wanted 2 X H2O = 1 given 0.038 mol X =
Step 2:
1.22 g oxygen
1 mole =
0.038 mol O2
32 g
O H2  H2O
given
wanted
wanted 2
X H2O = 1
given
0.038 mol
X =
0.076 mol H2O

11. = O2 + 2H2  2 H2O wanted 2 X H2O = 2 given 0.525 mol X =
Step 2:
1.05 g H2
1 mole =
0.525 mol H2
2 g
O H2  H2O
given
wanted
wanted 2
X H2O = 2
given
0.525 mol X =
0.525 mol H2O

12. The one that produces the smallest amount is your limiting reactant.
Step 3:
The one that produces the smallest amount is your limiting reactant.
1.22 g of O2 would produce mol H2O
Oxygen is your limiting reactant!
1.05 g of H2 would produce mol H2O

13. Limiting Reactants
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.

14. Step 1: Write a balanced equation.
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.
2Na Cl2  2NaCl

15. Step 2:
For each reactant, calculate the amount of product formed.
Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 L of chlorine gas at STP to produce sodium chloride.
2Na Cl2  2NaCl

16. = 2Na + Cl2  2NaCl wanted 2 X NaCl = 2 given 0.0739 mol X =
Step 2:
1.7 g Na
1 mole =
mol Na
23 g
2Na Cl2  2NaCl
wanted
given
wanted 2
X NaCl = 2
given
mol X =
mol NaCl

17. = 2Na + Cl2  2NaCl wanted 2 X NaCl = 1 given 0.116 mol X =
Step 2:
2.6 L Cl2
1 mole =
0.116 mol Cl2
22.4 L
2Na Cl2  2NaCl
wanted
given
wanted 2
X NaCl = 1
given
0.116 mol X =
0.232 mol NaCl

18. The one that produces the smallest amount is your limiting reactant.
*Step 3:
The one that produces the smallest amount is your limiting reactant.
1.7 g Na would produce mol NaCl
Sodium is your limiting reactant!
2.6 L Cl2 would produce mol NaCl

19. Percent Yield
measured in lab
calculated on paper

20. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical yield and % yield of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
? g
actual: 46.3 g

21. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
1 mol
K2CO3
138 g
2 mol
KCl
1 mol
K2CO3 74
g KCl
1 mol
KCl
= 49.1
g KCl

22. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.1 g
actual: 46.3 g
Theoretical Yield = 49.1 g KCl
46.3 g
49.1 g
% Yield =
 100 =
94.3%