1. Unit 6: Balancing Chemical Equations, Chemical Reactions & Stoichiometry
The
GowerHour

2. VII. Stoichiometric “road map” (Use the balanced chemical equation)
aA + bB cC + dD
Mol to Mol Ratio Using the Coefficients from the balanced chemical equation
Mass A
Mass B
Molar mass
Molar mass
Mol A
Mol B
6.022 x 1023
6.022 x 1023
Atoms
Molecules
Formula Units A
Atoms
Molecules
Formula Units B

3. atoms; particles; molecules; formula units
10.1
D. Converting Number of Particles to Moles
1. One mole (mol) of a substance is  1023 representative particles of that substance and is the SI unit for measuring the amount of a substance.
2. The number of representative particles in a mole,  1023, is called Avogadro’s number.
atoms; particles; molecules; formula units

4. 3. Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, tools, and garden furniture. How many moles of magnesium is 1.25 x 1023 atoms of magnesium?
1 mol Mg atoms = x 1023 atoms Mg
1 mol Mg atoms
1.25 x 1023 atoms Mg X
6.022 x 1023 atoms Mg
= mol Mg
= 2.08 x 10-1 mol Mg

5. atoms; particles; molecules; formula units
10.1
E. Converting Moles to Number of Particles
moles of methanol, CH3OH, contain how many molecules?
atoms; particles; molecules; formula units
1 mol molec CH3OH = x 1023 molec CH3OH
6.022 x 1023 molec CH3OH
1.56 mol CH3OH x
1 mol CH3OH
= x 1023 molec. CH3OH

6. Calculating Molecular Mass
Calculate the molecular mass of magnesium carbonate, MgCO3.
24.31 g g + 3(16.00 g) =
84.32 g/mol

7. 2. Aluminum satellite dishes are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass of 9.45 mol of aluminum oxide?
1 mol Al2O3 = g Al2O3
g Al2O3
9.45 mol Al2O3 x
= 964 g Al2O3
1 mol Al2O3
These aluminum satellite dishes at the National Radio Astronomy Observatory near Soccoro, New Mexico are naturally protected from corrosion by the formation of a thin film of aluminum oxide (Al2O3).

8. 4. When iron is exposed to air, it corrodes to form red-brown rust
4. When iron is exposed to air, it corrodes to form red-brown rust. Rust is iron(III) oxide (Fe2O3). How many moles of iron(III) oxide are contained in 92.2 g of pure Fe2O3?
1 mol Fe2O3 = g Fe2O3
1 mol Fe2O3
92.2 g Fe2O3 x
= mol Fe2O3
g Fe2O3
= x 10-1 mol Fe2O3
Rust weakens an iron chain.

9. Everything must go through Moles!!!
Calculations
molar mass Avogadro’s number Grams Moles particles
Everything must go through Moles!!!

10. mass
IV. Stoichiometry = The _____ relationships among reactants and products in chemical reactions. A chemical equation is a statement of ___________ fact. On the left side of the reaction are the _________, and on the right side the _________ of the reaction. Since no atoms are created nor destroyed in a ___________ chemical reaction, the equation must be _________. This means that the combined weight of the reactants is exactly ______ to the combined weight of the products. In terms of chemical laws, this is the Law of _____________ of matter (__________).
experimental
reactants
products
non-nuclear
balanced
equal
conservation
Lavoisier

11. V. Limiting and Excess Reagents
A. In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
B. The _____________ is the reagent that determines the amount of product that can be formed by a reaction.
C. In the reaction of nitrogen and hydrogen, hydrogen is the limiting reagent. Nitrogen is the reagent that is not completely used up in the reaction. The reagent that is not used up is called the ____________.
limiting reagent
excess reagent

12. D. Analogy: How to make a cheese sandwich.
  2 slices of bread + 1 slice of cheese →
1 cheese sandwich
  If you have 8 slices of bread and 6 slices of cheese, how many sandwiches can you make? __ (theoretical yield)
  What is the limiting reagent? ______
  What is the excess reagent? _______
  How much of the excess reagent is left at the end of the rxn? ______________ 4
bread
cheese
2 slices cheese

13. VI. Percent Yield
A. The percent yield is a measure of the __________ of a reaction carried out in the laboratory.
B. A batting average is actually a percent yield.
efficiency

14. theoretical
C. The ____________ yield is the maximum amount of product that could be formed from given amounts of reactants.
D. In contrast, the amount of product that actually forms when the reaction is carried out in the __________ is called the actual yield.
E. The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.
laboratory

15. VII. Stoichiometric “road map” (Use the balanced chemical equation)
aA + bB cC + dD
Mol to Mol Ratio Using the Coefficients from the balanced chemical equation
Mass A
Mass B
Molar mass
Molar mass
Mol A
Mol B
6.022 x 1023
6.022 x 1023
Atoms
Molecules
Formula Units A
Atoms
Molecules
Formula Units B

16. Examples:
A. Calcium reacts with oxygen to form calcium oxide. If you are given 80.0 g of calcium, (1) Calculate the theoretical yield of calcium oxide. (2) How many moles of your excess reagent (ER) are used? LR
Ca + O CaO 2 2 ER
1 mol Ca
2 mol CaO
56.08 g CaO
(1)
80.0 g Ca x x x
40.08 g Ca
2 mol Ca
1 mol CaO LR
= 112 g CaO
Theoretical Yield ER
1 mol Ca
1 mol O2
= mol O2
(2)
80.0 g Ca x x
40.08 g Ca
2 mol Ca LR
ER used

17. B. Nitrogen and oxygen react to form dinitrogen pentoxide
B. Nitrogen and oxygen react to form dinitrogen pentoxide. If 112 g of nitrogen react with unlimited oxygen, (1) how many moles of the excess reagent (ER) react and (2) how many grams of dinitrogen pentoxide (theoretical yield) are formed?
(112 g) 2
N O N2O5 5 2 LR ER
1 mol N2
5 mol O2
(1)
112 g N2 x x
= 9.99 mol O2
28.02 g N2
2 mol N2 LR
ER used
1 mol N2
2 mol N2O5
g N2O5
(2)
112 g N2 x x x
28.02 g
2 mol N2
1 mol N2O5 LR
= 432 g N2O5
Theoretical Yield

18. C. Aqueous sodium chloride and aqueous silver nitrate react forming solid silver chloride (a precipitate) and aqueous sodium nitrate. If one solution contains g of silver nitrate and it reacts with unlimited sodium chloride, how many grams of silver chloride (Theoretical Yield) can be made? ER LR
NaCl (aq) + AgNO3 (aq) AgCl (s) + NaNO3 (aq) LR
1 mol AgNO3
1 mol AgCl g
100.0 g AgNO3 x x x g
1 mol AgNO3
1 mol AgCl
= g AgCl
Theoretical Yield

19. D. How many (1) grams (theoretical yield) and (2) molecules of carbon dioxide gas are produced when g of methane gas (CH4) are burned. Methane reacts with oxygen to form carbon dioxide and water. LR
CH O2  CO H2O 2 2 ER
1 mol CH4
1 mol CO2
44.01 g CO2
(1)
100.0 g CH4 x x x
16.05 g
1 mol CH4
1 mol CO2 LR
= g CO2
Theoretical Yield
1 mol CH4
1 mol CO2
6.022 x 1023 molec CO2
(2)
100.0 g CH4 x x x
16.05 g
1 mol CH4
1 mol CO2 LR
= x 1024 molec CO2

20. E. Nitrogen reacts with hydrogen gas forming ammonia (nitrogen trihydride). How many (1) grams of hydrogen and (2) moles of nitrogen are required to form 1.00 pound of ammonia?
N H2  NH3 3 2
453.6 g NH3
1 mol NH3
3 mol H2
2.02 g H2
(1)
1.00 lb NH3 x x x x
1 lb
17.04 g
2 mol NH3
1 mol H2
= 80.7 g H2
453.6 g NH3
1 mol NH3
1 mol N2
(2)
1.00 lb NH3 x x x
= 13.3 mol N2
1 lb
17.04 g
2 mol NH3

21. Review Question:
Nitrogen and hydrogen gas combine to form hydrazine
(dinitrogen tetrahydride) which is used for rocket fuel.
How many grams of nitrogen gas are needed to form 155
g of hydrazine?
N H2  N2H4 x
155 g N2H4
32.06 g
1 mol N2H4
1 mol N2
28.02 g N2
= 135 g N2

22. F. How much hydrogen peroxide can be formed (theoretical yield) from 10.0 g of hydrogen and 125 g of oxygen? What is the limiting reagent?
H2 + O2  H2O2
1 mol H2
1 mol H2O2
34.02 g H2O2
10.0 g H2 x x x
= 168 g H2O2
2.02 g
1 mol H2
1 mol H2O2
1 mol O2
1 mol H2O2
34.02 g H2O2
125 g O2 x x x
= 133 g H2O2
32.00 g
1 mol O2
1 mol H2O2
L.R. = O2
Theoretical yield

23. G. Determine the theoretical yield of water that can be formed from 10
G. Determine the theoretical yield of water that can be formed from 10.0 g of hydrogen and g of oxygen. What is the limiting reagent? How much excess reagent is remaining? ER 2
H2 + O2  H2O 2
1 mol H2
1 mol H2O
18.02 g H2O
10.0 g H2 x x x
= 89.2 g H2O
2.02 g
1 mol H2
1 mol H2O
L.R. = H2
Theoretical yield
1 mol O2
2 mol H2O
18.02 g H2O
100.0 g O2 x x x
= g H2O
32.00 g
1 mol O2
1 mol H2O
1 mol H2
1 mol O2
32.00 g O2
10.0 g H2 x x x
= 79.2 g O2
2.02 g
2 mol H2
1 mol O2
ER used
100.0 g O g O2 = 20.8 g O2 remaining

24. H. How many grams of aluminum oxide can be formed (theoretical yield) from g of aluminum and g of oxygen gas? How much excess reagent is remaining? 4
Al O2  Al2O3 3 2
1 mol Al
2 mol Al2O3 g
100.0 g Al x x x
= g Al2O3
26.98 g
4 mol Al
1 mol Al2O3
L.R. = Al
Theoretical yield
1 mol O2
2 mol Al2O3 g
100.0 g O2 x x x
= g Al2O3
32.00 g
3 mol O2
1 mol Al2O3
1 mol Al
3 mol O2
32.00 g O2
100.0 g Al x x x
= g O2
26.98 g
4 mol Al
1 mol O2
100.0 g O g O2 = g O2 Remaining

25. I. A student does a lab in which she makes 2.17 g of Lead (II) nitrate. From her limiting reagent she calculated that she should have made 2.55 g. Determine: (1) theoretical yield, (2) actual yield, (3) % yield (4) absolute and relative error.
(1) g Pb(NO3)2
(2) g Pb(NO3)2
(4) Ea = g – 2.17 g = g
Note: % yield + Er = 100 %

26. J. A sample of 100. 0 g of hydrazine (N2H4) burns in 280
J. A sample of g of hydrazine (N2H4) burns in g of oxygen gas and 97.2 g of water are formed. Calculate the % yield of water. (Nitrogen dioxide is also produced.) Calculate the % error (relative error).
N2H O2  H2O NO2 3 2 2
1 mol N2H4
2 mol H2O
18.02 g
100.0 g N2H4 x x x
= g H2O
32.06 g
1 mol N2H4
1 mol H2O
1 mol O2
2 mol H2O
18.02 g
280.0 g O2 x x x
= g H2O
32.00 g
3 mol O2
1 mol H2O LR
Theo. yield
% error = 100% % = 7.5%

27. 2 AgNO3 + CaCl2  Ca(NO3)2 + 2 AgCl (s)
K. In lab a student mixes a solution that contains 1.70 g silver nitrate and a solution that contains g calcium chloride. Aqueous calcium nitrate and solid silver chloride are produced. After washing, filtering, and drying the precipitate, he finds that he has obtained 1.25 g of silver chloride. What is the % yield?
2 AgNO3 + CaCl2  Ca(NO3) AgCl (s)
1.70 g AgNO3 g
1 mol AgNO3
2 mol AgNO3
2 mol AgCl
1 mol AgCl g x
= 1.43 g AgCl
Theo. yield
0.832 g CaCl2 g
1 mol CaCl2
2 mol AgCl
1 mol AgCl g x
= 2.15 g AgCl

28. IQ1
Determine the mass of Iron III oxide formed when 25.1 g of iron reacts with oxygen.
Balanced equations: ______________________________
2. What mass of oxygen is needed to react with 6.8 x 1023 atoms of iron? (Use the above equations).
4 Fe + 3 O2 → 2 Fe2O3
25.1 g Fe
2 mol Fe2O3
g Fe2O3
1 mol Fe
55.85 g Fe
4 mol Fe
1 mol Fe2O3
= 35.9 g Fe2O3
6.8 x 1023 atoms Fe
1 mol Fe
3 mol O2
32.00 g O2
6.022 x 1023 atoms Fe
4 mol Fe
1 mol O2
= 27 g O2

29. IQ 2
Determine the (a) limiting reagent and (b) theoretical yield of iron when 82.3 g of iron (III) oxide reacts with 79.3 g of carbon monoxide. (c) Grams of excess reagent remaining.
Balance equ.: __________________________________
(a)
(b)
Grams excess reagent remaining?
Fe2O CO → Fe + CO2 3 2 3
82.3 g Fe2O3
1 mol Fe2O3
3 mol CO
28.01 g CO
g Fe2O3
1 mol Fe2O3
1 mol CO
Fe2O3 – Limiting Reagent
43.3 g CO needed
82.3 g Fe2O3
1 mol Fe2O3
2 mol Fe
55.85 g Fe
g Fe2O3
1 mol Fe2O3
1 mol Fe
57.6 g Fe
Excess = start – grams reacted
Excess = 79.3 g CO – 43.3 g CO
= 36.0 g CO

30. Stoichiometry WS I
a mol O2
b g KCl
a g H2O
b g CO2
a mol FeCl3
b mol Cu

31. Stoichiometry WS II
a mol O2
b g KCl
a g H2O
b g CO2
a mol FeCl3
b mol Cu

32. Sample Test Part A
(1) Syn (2) Decomp (3) DD (4) SD (5) Combustion
a. 1, 4, 3, 4 b. 1, 1, 1, 1
c. 6, 1, 2, 3 d. 2, 3, 1, 3
a. 4, 3, 2 (Syn) b. 2, 2, 2, 1 (SD)
c. 2, 7, 4, 6 (Comb)
d. 1, 2, 1, 2 (DD) e. 2, 1, 1 (Decomp)
a. 1, 1, 2 AgBr, 1 CaSO4 b. 2, 1 H2, 1 Br2
c. 1, 5, 3 CO2, 4 H2O d. 2, 3, 1 Al2S3
e. 2, 1 N2, 3 H2

33. Sample Test Part A
a. 2, 3, 1 Al2(SO4)3 Cu
b. 1, 2, 1 Ca(OH)2, 1H2
c. 1, 2, 1 Fe(OH)2, 1 H2
d. 1, 2, 1 ZnCl2, 1 H2
e. NR
f. 1, 2, 1 Cu(NO3)2, 2 Ag
HOFBrINCl
Speeds up rxn w/o being used up in the rxn.
C, CO, CO2, H2O
See Metal activities series.

34. Sample Test Part B
a. Fe b. O2
a x 1023 molec b. 138 g NaCl
c x 1023 molec Cl2
a. AgNO3 b g Ag c. 89 g Cu
74.3 %
When LR runs out, the rxn stops.
145 g NH3
F, A , E, C , D, B

35. Unit 6 HW Packet Schedule Unit 6 Study Guide Stoichiometry WS I
Stoichiometry WS II
Limiting Reagent & Percent Yield WS
Theoretical Yield/Percent Yield WS
Reactions WS I
Reactions WS II
Reactions WS III
Stoichiometry Sample Test

36. Chemistry WS
a mol O2
b g KCl
a g H2O
b g CO2
a mol FeCl3
b mol Cu

37. Stoichiometry WS 1.81 x 1023 molecules H2 7.3 g H2 3.09 mol Al2(SO4)3
22 g Al
6.7 x 1022 atoms Al
8.1 g H2

38. Limiting Reagent & Percent Yield WS
a. L.R. = HCl
b. L.R. = O2
c. L.R. = Cl2
a. L.R. = HCl; 0.26 mol H2 produced
b mol Zn used; 0.04 mol Zn excess
a. L.R. = HCl; 0.12 mol AlCl3 produced
b mol Al used; 0.03 mol Al excess

39. Limiting Reagent & Percent Yield
87.9 %
75.2 %
Theoretical Yield/Percent Yield
54.2 %
Actual yield = 1660 g NH3
37.0 %
Actual yield = 58.7 g MgO

40. Reaction Types
DD: Al2(SO4)3 + Ca3(PO4)2  2 AlPO4 + 3 CaSO4
DD: MgCl2 + 2 AgNO3  Mg(NO3)2 + 2 AgCl
Syn: 2 H2 + O2  2 H2O
SD: Zn + Cu(NO3)2  Zn(NO3)2 + Cu
DD: CuO + H2SO4  CuSO4 + H2O
SD: 2 NaI + Cl2  2 NaCl + I2
Decomp: Cu(OH)2  CuO + H2O
12. SD: K + H2O  KOH + H2

41. Sample Test Part A
(1) Syn (2) Decomp (3) DD (4) SD (5) Combustion
a. 1, 4, 3, 4 b. 1, 1, 1, 1
c. 6, 1, 2, 3 d. 2, 3, 1, 3
a. 4, 3, 2 (Syn) b. 2, 2, 2, 1 (SD)
c. 2, 7, 4, 6 (Comb)
d. 1, 2, 1, 2 (DD) e. 2, 1, 1 (Decomp)
a. 1, 1, 2 AgBr, 1 CaSO4 b. 2, 1 H2, 1 Br2
c. 1, 5, 3 CO2, 4 H2O d. 2, 3, 1 Al2S3
e. 2, 1 N2, 3 H2

42. Sample Test Part A
a. 2, 3, 1 Al2(SO4)3 Cu
b. 1, 2, 1 Ca(OH)2, 1H2
c. 1, 2, 1 Fe(OH)2, 1 H2
d. 1, 2, 1 ZnCl2, 1 H2
e. NR
f. 1, 2, 1 Cu(NO3)2, 2 Ag
HOFBrINCl
Speeds up rxn w/o being used up in the rxn.
C, CO, CO2, H2O
See Metal activities series.

43. Sample Test Part B
a. Fe b. O2
a x 1023 molec b. 138 g NaCl
c x 1023 molec Cl2
a. AgNO3 b g Ag c. 89 g Cu
74.3 %
When LR runs out, the rxn stops.
145 g NH3
F, A , E, C , D, B

44. Stoichiometry Review WS
a. N2O5 + H2O  2 HNO b g HNO3
c g H2O
2. a. 4 Al + 3 PbO2  2 Al2O3 + 3 Pb b g Al
3. a. LR = H2S b g (NH4)2S c g NH3
4. a. Yes; SD; Cl2 + 2KBr  2 KCl + Br2
b. LR = Cl2 c g Br2 d g KBr
5. a. LR = As2O3 b %
6. a. 6, 6, 1, 6 b x 103 g O2
7. a. 71 % b. 26 metric tons

45. Unit 6 HW Packet Schedule Unit 6 Study Guide Stoichiometry WS I
Stoichiometry WS II
Limiting Reagent & Percent Yield WS
Theoretical Yield/Percent Yield WS
Reactions WS I
Reactions WS II
Reactions WS III
Stoichiometry Sample Test