1. Section 6.5—Stoichiometry
How can we determine in a lab the concentration of electrolytes?

2. What do those coefficients really mean?
For every 2 moles of H2…
and 2 moles of H2O are produced 2 2
2 H2 + O2  2 H2O
No coefficient = 1
1 mole of O2 is need to react…

3. What is stoichiometry?
Stoichiometry – Using the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the equation.

4. Stoichiometry with Moles
Example:
If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed?
2 H2 + O2  2 H2O

5. Stoichiometry with Moles
Example:
If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed?
2 H2 + O2  2 H2O
From balanced equation:
2 mole H2  1 mole O2
4.2 mole H2 1
mole O2
= ________ mole O2
2.1 2
mole H2

6. But we can’t measure moles in lab!
We can’t go to the lab and count or measure moles…so we need a way to work in measurable units, such as grams and liters!
Molecular mass gives the grams = 1 mole of a compound!

7. Stoichiometry with Moles & Mass
Example:
How many grams of AgCl will be precipitated if
0.45 mole AgNO3 is reacted as follows:
2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2

8. Stoichiometry with Moles & Mass
Example:
How many grams of AgCl will be precipitated if
0.45 mole AgNO3 is reacted as follows:
2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2
From balanced equation:
2 mole AgNO3  2 mole AgCl
Molar Mass of AgCl:
1 mole AgCl = g
0.45 mole AgNO3 2
mole AgCl
143.35
g AgCl
= ________ g AgCl 65 2
mole AgNO3 1
mole AgCl

9. Stoichiometry with Mass
Example:
How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction:
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

10. Stoichiometry with Mass
Example:
How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction:
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Molar Mass of NaOH:
1 mole NaCl = g
From balanced equation:
2 mole NaOH  1 mole Ba(OH)2
Molar Mass of Ba(OH)2:
1 mole Ba(OH)2 = g
14.5 g NaOH 1
mole NaOH 1
mole Ba(OH)2
171.35
g Ba(OH)2
40.00
g NaOH 2
mole NaOH 1
mole Ba(OH)2
= ________ g Ba(OH)2
31.1

11. But what about for solutions?
Molarity gives the number of moles of the solute that are in 1 L of a solution

12. Stoichiometry with Solutions
Example:
If you need 15.7 g Ba(OH)2 to precipitate, how many liters of 2.5 M NaOH solution is needed?
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

13. Stoichiometry with Solutions
Example:
If you need 15.7 g Ba(OH)2 to precipitate, how many liters of 2.5 M NaOH solution is needed?
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Concentration of NaOH:
2.5 mole NaOH = 1 L
From balanced equation:
2 mole NaOH  1 mole Ba(OH)2
Molar Mass of Ba(OH)2:
1 mole Ba(OH)2 = g
15.7 g Ba(OH)2 1
mole Ba(OH)2 2
mole NaOH 1
L NaOH
171.35
g Ba(OH)2 1
mole Ba(OH)2
2.5
mole NaOH
= ________ L NaOH
0.0733

14. What about gases?
Standard Temperature and Pressure (STP) – 1 atm (or kPa) and 273 K (0°C)
Molar Volume of a Gas – at STP, 1 mole of any gas = 22.4 liters

15. Stoichiometry with Gases
Example:
If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)

16. Stoichiometry with Gases
Example:
If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)
Molar volume of a gas:
1 mole H2 = 22.4 L
From balanced equation:
1 mole Zn  1 mole H2
Molar Mass of Zn:
1 mole Zn = g
1.5 g Zn 1
mole Zn 1
mole H2
22.4
L H2
65.39
g Zn 1
mole Zn 1
mole H2
= ________ L H2
0.51

17. Keeping all these equalities straight!
TO GO BETWEEN
USE THE EQUALITY
Grams & moles
Molecular Mass in grams = 1 mole
moles & liters of a solution
Molarity in moles = 1 L
Moles & liters of a gas at STP
1 mole = 22.4 L at STP
2 different chemicals in a reaction
Coefficient ratio from balanced equation

18. Titrations—Using Stoichiometry
Titration – Addition of a known volume of a known concentration solution to a known volume of unknown concentration solution to determine the concentration.

19. End Point
End Point (or Stoichiometric Point) – When there is no reactant left over—they have all be reacted and the solution contains only products
The end point must be reached in order to use stoichiometry to calculate the unknown solution concentration
Indicators – Paper or liquid that change color based on pH level.
If the pH of the products is known, the indicator can be chosen to indicate the end point

20. Gravemetrics—Using Stoichiometry
Gravemetric Analysis – Using a reaction to precipitate out an insoluble compound. The solid is dried and massed. Stoichiometry can then be used to determine the original substance’s concentration from the mass of the precipitate

21. If you are making 0.57 moles H2O, how many moles of O2 are needed?
Let’s Practice #1
Example:
If you are making 0.57 moles H2O, how many moles of O2 are needed?
2 H2 + O2  2 H2O

22. If you are making 0.57 moles H2O, how many moles of O2 are needed?
Let’s Practice #1
Example:
If you are making 0.57 moles H2O, how many moles of O2 are needed?
2 H2 + O2  2 H2O
From balanced equation:
2 mole H2O  1 mole O2
0.57 mole H2O 1
mole O2
= ________ mole O2
0.29 2
mole H2O

23. 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Let’s Practice #2
Example:
If you need to precipitate 10.7 g of Ba(OH)2, how many grams NaOH are needed?
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

24. 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Let’s Practice #2
Example:
If you need to precipitate 10.7 g of Ba(OH)2, how many grams NaOH are needed?
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Molar Mass of Ba(OH)2:
1 mole Ba(OH)2 = g
From balanced equation:
2 mole NaOH  1 mole Ba(OH)2
Molar Mass of NaOH:
1 mole NaCl = g
10.7 g Ba(OH)2 1
mole Ba(OH)2 2
mole NaOH
40.00
g NaOH
171.35
g Ba(OH)2 1
mole Ba(OH)2 1
mole NaOH
= ________ g NaOH
5.00

25. Let’s Practice #3 Example:
How many moles AgNO3 are needed to react with 10.7 g CaCl2?
2 AgNO3 + CaCl2  2 AgCl + 2 Ca(NO3)2

26. Let’s Practice #3 Example:
How many moles AgNO3 are needed to react with 10.7 g CaCl2?
2 AgNO3 + CaCl2  2 AgCl + 2 Ca(NO3)2
From balanced equation:
2 mole AgNO3  1 mole CaCl2
Molar Mass of CaCl2:
1 mole CaCl2 = g
10.7 g CaCl2 1
mole CaCl2 2
mole AgNO3
= ______ mole AgNO3
0.193
110.98
g CaCl2 1
mole CaCl2

27. 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Let’s Practice #4
Example:
How many liters of 0.10 M NaOH is needed to react with L of 0.25 M BaCl2?
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

28. 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Let’s Practice #4
Example:
How many liters of 0.10 M NaOH is needed to react with L of 0.25 M BaCl2?
2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl
Concentration of NaOH:
0.10 mole NaOH = 1 L
From balanced equation:
2 mole NaOH  1 mole BaCl2
Concentration of BaCl2:
0.25 mole BaCl2 = 1 L BaCl2
0.125 L BaCl2
0.25
mole BaCl2 2
mole NaOH 1
L NaOH 1
L BaCl2 1
mole BaCl2
0.10
mole NaOH
= ________ L NaOH
0.625

29. 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)
Let’s Practice #5
Example:
If you produce 15.4 L of H2 at STP, how many grams of ZnCl2 is also produced?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)

30. 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)
Let’s Practice #5
Example:
If you produce 15.4 L of H2 at STP, how many grams of ZnCl2 is also produced?
2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)
Molar Mass of ZnCl2:
1 mole ZnCl2 = g
From balanced equation:
1 mole ZnCl2  1 mole H2
Molar volume of a gas:
1 mole H2 = 22.4 L
15.4 L H2 1
mole H2 1
mole ZnCl2
136.29
g ZnCl2
22.4
L H2 1
mole H2 1
mole ZnCl2
= ________ g ZnCl2
93.7