Osterberg-Chemistry
Ch. 7/9-Moles and Stoichiometry
Name _____________________________
Date ___________________ Hour______
Chapter 7/9 Objectives-Part II
Obj.1Stoichiometry
• Stoichiometry: mass and amount relationships between reactants & products in a
chemical reaction.
• Rules and Hints for Solving Stoichiometry Problems
1. Start with a balanced equation.
2. A ↔ the substance you start with on top (the number is given)
3. B ↔ the substance you end with (what the problem is asking for)
4. Use conversion factors to cancel out the units and move onto the next unit.
• Conversion factors are a form of one
1 hour = 60 seconds
• Examples:
1 mile = 5280 feet
⎛ 1 hr ⎞
⎜
⎟ or
⎝ 60 sec ⎠
⎛ 1 mi ⎞
⎜
⎟ or
⎝ 5280 ft ⎠
⎛ 60 sec ⎞
⎜
⎟
⎝ 1 hr ⎠
⎛ 5280 ft ⎞
⎜
⎟
⎝ 1 mi ⎠
5. Multiply across (top and bottom), then divide the top number by the bottom to get
your answer. Round to the correct umber of significant figures.
• Problem Example: How many minutes are there in 12.000 years?
A ↔ 12.000 years B ↔ ? seconds
12.000 years ⎛ 365.25 days ⎞ ⎛ 24 hours ⎞
⎜
⎟⎜
⎟
⎝ 1 year ⎠ ⎝ 1 day ⎠
⎛ 60 min ⎞
⎜
⎟ = 6,311,500 min
⎝ 1 hour ⎠
Obj. 2Conversion Factors for Stoichiometry
Moles
(The Bridge)
Mass
Volume
Conversion factor
# mole A = # mole B
1 mole A= # grams A
1 mole B = # grams B
1 mole A = 22.4 L A
1 mole B = 22.4 L B
Representative 1 mole A = 6.02 x 1023
Particles
particles of A
1 mole B = 6.02 x 1023
particles of B
Helpful Hints
Plug in the numbers from the coefficients in
the balanced equation in front of
substances A and B
Plug in the molar mass of A or B (To find
the molar mass, add up all the elements in
the compound.)
The unit for volume is Liters (L).
Representative particles could be atoms,
molecules, or formula units depending on
what A or B is.
Osterberg-Chemistry
Ch. 7/9-Moles and Stoichiometry
Name _____________________________
Date ___________________ Hour______
Obj. 3Stoichiometry Examples
4NH3 + 5O2 → 4NO + 6H2O
• Ex1 (Mole A to Mole B): How many moles of H2O are produced if you have 3
moles of NH3?
A ↔ 3 moles NH 3 B ↔ ? moles H 2O
3 mol NH 3 ⎛ 6 mol H 2O ⎞ 18
= 4.5 ≈ 5 mol H 2O
⎜
⎟=
4
mol
NH
4
⎝
3⎠
• Ex2 (Mole A to Mass B): How many g of H2O are produced if you have 3 moles of
NH3?
A ↔ 3 moles NH 3 B ↔ ? g H 2O
3 mol NH 3 ⎛ 6 mol H 2O ⎞⎛18.02 g H 2O ⎞ 324.36
= 81.09 ≈ 80 g H 2O
⎜
⎟⎜
⎟=
4
⎝ 4 mol NH 3 ⎠⎝ 1 mol H 2O ⎠
• Ex3 (Mass A to Mass B): If you need 16.32 g of O2 with ammonia, how many g of
NO were produced?
A ↔ 16.32 g O 2 B ↔ ? g NO
16.32 g O 2 ⎛ 1 mol O 2 ⎞⎛ 4 mol NO ⎞⎛ 30.01 g NO ⎞ 1959.0528
= 12.24408 ≈ 12.24 g NO
⎟=
⎜
⎟⎜
⎟⎜
⎝
⎠
32.00
g
O
5
mol
O
1
mol
NO
160
⎝
2 ⎠⎝
2 ⎠
• Ex4 (Moles A to Volume B): If you have 8 moles NH3, how many L of O2 gas are
needed?
A ↔ 8 mol NH 3 B ↔ ? L O 2
8 mol NH 3 ⎛ 5 mol O 2 ⎞⎛ 22.4 L O 2 ⎞ 896
= 224 ≈ 200 L O 2
⎜
⎟⎜
⎟=
4
⎝ 4 mol NH 3 ⎠⎝ 1 mol O 2 ⎠
• Ex5 (Mass A to Rep. Part. of B): How many molecules of H2O were produced if
you used 16.32 g of O2?
A ↔ 16.32 g O 2 B ↔ ? molecules H 2O
16.32 g O 2 ⎛ 1 mol O 2 ⎞⎛ 6 mol H 2O ⎞⎛ 6.02 × 10 23 molec H 2O ⎞ 5.894784 × 10 25
=
⎜
⎟⎜
⎟⎜
⎟=
1 mol H 2O
160
⎝ 32.00 g O 2 ⎠⎝ 5 mol O 2 ⎠⎝
⎠
3.68424 × 10 23 ≈ 3.684 × 10 23 molecules H 2O
Osterberg-Chemistry
Ch. 7/9-Moles and Stoichiometry
Name _____________________________
Date ___________________ Hour______
Obj. 4Limiting/ Excess Reactant
• Excess reactant: reactant that is not completely used up.
• Limiting reactant: reactant that gets used up first.
• Hints for Limiting Reactant Problems
1. Write balanced reaction.
2. Find the limiting reactant by starting with each reactant and choose one product
for the B (only need to go to moles B).
o A’s are the two reactants (Two stoichiometry problems)
o B is moles of the one product you choose (Go to the same product for both
A’s)
3. Find the number of moles (or grams) of excess reactant by starting with how
much extra product could be made by subtracting your answers from step 2 and
going to moles (or grams) of your excess reactant.
o A is found by subtracting your two answers from step 2.
o B is the moles (or grams) of your excess reactant
4. Find the number of moles (or grams) of all products formed based on limiting
reactant from step 2 b/c that is what runs out first.
o Copy the limiting reactant answer from step 2.
o A is the limiting reactant.
o B is the moles (or grams) of the other product you did not go to in step 2.
• Ex: The reaction starts with 44.0 g of MgSO4 and 48.2 g of KCl.
MgSO4 + 2 KCl  MgCl2 + K2SO4
a) What is the limiting reactant?
A1 ↔ 52.0 g MgSO 4 A2 ↔ 8.21 g KCl B ↔ ? mol MgCl2
44.0 g MgSO 4 ⎛ 1 mol MgSO 4 ⎞ ⎛ 1 mol MgCl2 ⎞
44.0
= 0.365539 ≈ 0.366 mol MgCl2
⎜
⎟⎜
⎟=
⎝ 120.37 g MgSO 4 ⎠ ⎝ 1 mol MgSO 4 ⎠ 120.37
48.2 g KCl ⎛ 1 mol KCl ⎞ ⎛ 1 mol MgCl2 ⎞ 48.2
=
= 0.32327 ≈ 0.323 mol MgCl2
⎜
⎟
⎝ 74.55 g KCl⎠ ⎝ 2 mol KCl ⎠ 149.1
Therefore, KCl is the limiting reactant.
b) How many extra moles are there of the excess reactant?
A ↔ 0.366 − 0.323 = 0.043 mol MgCl2
B ↔ ? mol MgSO 4
0.043 mol MgCl2 ⎛ 1 mol MgSO 4 ⎞
⎜
⎟ = 0.043 mol MgSO 4
⎝ 1 mol MgCl2 ⎠
c) How many moles of each of the products will be formed?
0.323 mol MgCl2
48.2 g KCl ⎛ 1 mol KCl ⎞ ⎛ 1 mol K 2SO 4 ⎞
= 0.323 mol MgSO 4
⎜
⎟
⎝ 74.55 g KCl⎠ ⎝ 2 mol KCl ⎠