1. Chapter 9

2. Stoichiometry
The calculation of quantities in chemical reactions

3. Particles
N2(g) + 3H2(g) NH3(g)
1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3
Ratio : 1:3:2 particles

4. Moles 1 mole = 6.02 x 1023 representative particles
Based on the reaction – mole ratio is 1:3:2
* Coefficients are the relative # of moles

5. Mass Law of conservation of mass 1:3 2 28g + 6 g 34 g1:
28g + 6 g g
N = 2x14 H = 2x3 NH3 = 17x2

6. Volume
Assume STP
22.4 L/mol

7. Example
2H2S + 3O SO2 + 2H2O

8. Mass of Reactants 2 mol x 34.1 g = 68.2g mol 3 mol x 32 g = 96g

9. Mass of Products 2 mol x 64.1g = 128.2 g mol 2 mol x 18g = 36 g

10. Volume of gases at STP Reactants 2 mol H2S x 22.4 L = 44.8 L H2S 1 mol
3 mol O2 x L = L O2

11. Products
2 mol SO2 x L = L SO2
1 mol
2 mol H2O x L = L H2O

12. Mole to Mole Calculations
How many moles of ammonia are produced when 0.06 mol of nitrogen reacts with hydrogen?

13. N2 + 3H NH3
0.06 mol N2 x 2mol NH3
1 mol N2
= .12mol NH3

14. Example Aluminum Oxide is formed from aluminum and oxygen
a) Write a balanced equation
4Al + 3O Al2O3

15. b) How many moles of Al are needed to form 2.3 mol of Al2O3 ?
2.3 mol Al2O3 x 4 mol Al
2 mol Al2O3
= 4.6mol Al

16. c) How many moles of Oxygen are required to react completely with 0
c) How many moles of Oxygen are required to react completely with 0.84 mol of Al ?
.84mol Al x 3mol O2
4mol Al
= .63 mol O2

17. d) Calculate the # of moles of Al2O3 formed when 17
d) Calculate the # of moles of Al2O3 formed when 17.2 mol of O2 reacts with Al ?
17.2 mol O2 x 2mol Al2O3
3mol O2
= mol Al2O3

18. Mass – Mass Calculations
GFM
Ex) H2
1 mol or 2g
2g mol

19. Example
Calculate the number of grams of ammonia produced by the reaction of 5.40 g of hydrogen with nitrogen.
N2(g) + 3H2(g) NH3(g)

20. 5.40g H2 x 1mol H2 x 2mol NH3 x 17g NH3
2g H mol H mol NH3
= 30.6g NH3
These problems are solved the same way as mole – mole problems

21. Steps Change mass to moles G (given) Change moles of G to moles of W
Change moles of W to grams of W

22. Example a) How many grams of O2 are required to burn 13g of C2H2 ?
2C2H2 + 5O CO2 + 2H2O
a) How many grams of O2 are required to burn 13g of C2H2 ?
= 40.0g O2

23. b) How many grams of CO2 and H2O are produced when 13g of C2H2 react with the oxygen.
13g C2H2 x 1mol C2H2 x 4mol CO2 x 44g CO2
26g C2H mol C2H mol CO2
= 44g CO2

24. Continued 13g C2H2 x 1mol C2H2 x 2mol H2O x 18g H2O
26g C2H mol C2H2 1mol H2O
= 9g H2O

25. Other Stoichiometric Calculations
Mass – Volume
Volume – Volume
Particle – Mass

26. Example
Tin plus hydroflouric acid yields Tin(II)flouride and hydrogen gas
Sn(s) + 2HF(g) SnF2(s) + H2(g)

27. a) How many grams of SnF2 can be
made by reacting 7.42 x 1024 molecules
of HF with tin?
966g SnF2

28. b) How many L of H2 (STP) are produced by reacting 23.4g of Sn with HF?
4.42 L H2

29. c) How many L of HF are needed to produce 14.2 L of H2(STP) ?
28.4 L HF

30. d) How many molecules of H2 are produced by the reaction of Sn with 80 L of HF (STP) ?
1.08 x 1024 molecules H2

31. Limiting Reagent Example
1lb of swiss & 1lb of pastrami and 14 slices of bread
You can only make 7 sandwiches
Limiting reagent - bread
Excess reagent – luncheon meats

32. Example
Sodium Chloride is prepared by the reaction of sodium metal with chlorine gas
2Na(s) + Cl2(g) NaCl(s)
What will occur when 6.70mol of Na reacts with 3.20 mol of Cl2 ?

33. a) What is the limiting reagent ?
6.70mol Na x 1mol Cl2
2mol Na
= 3.35 mol Cl2 required
* 3.35 mol Cl2 are required to completely react with 6.70mol Na but you only have 3.20 mol Cl2.
Cl2 is the limiting reagent.

34. b) How many moles of NaCl are produced?
3.20mol Cl2 x 2mol NaCl
1mol Cl2
= 6.40 mol NaCl

35. 3.20mol Cl2 x 2mol Na 1mol Cl2 = 6.40mol Na 6.70mol Na - 6.40mol Na
c) How much of the excess reagent remains un-reacted ?
3.20mol Cl2 x 2mol Na mol Cl2
= 6.40mol Na
6.70mol Na
- 6.40mol Na
.30mol Na is excess

36. Example
C2H4 + 3O CO2 + 2H2O
If 2.70 mol C2H4 reacts with 6.30 mol 02
a) What is the limiting reagent?

37. the limiting reagent is O2
2.7mol C2H4 x 3mol O2 = 8.1mol O2
1mol C2H4
* You need 8.1mol O2 to react with 2.7 mol C2H4
the limiting reagent is O2

38. 6.3 mol O2 x 2mol H2O 3mol O2 = 4.2mol H2O
b) Calculate moles of H2O produced
6.3 mol O2 x 2mol H2O
3mol O2
= 4.2mol H2O

39. c) Calculate moles of excess reagent
6.30mol O2 x 1mol C2H4
3mol O2
= 2.1mol C2H4

40. Final Step
2.7mol C2H4
- 2.1mol C2H4
0.60mol C2H4

41. Example
a) How many grams of H2 can be produced when 4g HCl is added to 3g of Mg?

42. 3g Mg x 1mol Mg = .12mol Mg 24.3g Mg .11mol HCl x 1mol H2 x 2g H2
4g HCl x 1mol HCl = mol HCl
36g HCl
3g Mg x 1mol Mg = .12mol Mg
24.3g Mg
.11mol HCl x 1mol H2 x 2g H2
2mol HCl mol H2
=0.11g H2

43. b) Assuming STP, what is the volume of H2?
0.11g H2 x 1mol x L
2g mol
= 1.23 L H2

44. Percent Yield
Theoretical Yield – the maximum amount of product that could be formed from a given amount of reactant

45. Actual Yield – Amount of product that forms when the reaction is carried out in the lab.

46. Percent Yield – Ratio of actual yield to theoretical yield
% Yield = actual yield x 100
theoretical yield

47. Example
CaCO CaO + CO2
*Calculate the % yield of CaO if 24.89g CaCO3 is heated to give 13.1g of CaO

48. 24.8g CaCO3 x 1mol CaCO x
100.1g CaCO3
1mol CaO x g CaO
1mol CaCO mol CaO
= 13.9g CaO

49. Percent Yield
13.1
13.9
94.2% x
100 =

50. Example
What is the % yield of copper if 3.74g of copper is produced when 1.87g of Aluminum is reacted with an excess of copper(II)sulfate?

51. 2Al + 3CuSO4 Al2(SO4)3 + 3Cu 1.87g Al x 1mol Al x 3mol Cu x 63.54g
*3.74g Cu
1.87g Al x 1mol Al x 3mol Cu x g
26.98g Al 2mol Al mol Cu
= 6.61g Cu

52. Percent Yield
3.74g Cu x =
6.61g Cu
56.6 %