1

2
 Determine mole ratios from coefficients in a reaction
 Predict quantities of reactants and products in chemical reactions
 Relate moles and gas volumes by stoichiometry (Ch.
13.3)
 Determine limiting reactants in a chemical reaction
 Determine excess reactant and how much INXS
 Determine theoretical, actual and percentage yields

3
Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction
It is based on the chemical equation
(Ch. 9) and the relationship between mass and moles (Ch. 10).

We can interpret a chemical equation in terms of number of molecules (or ions or formula units) or in terms of number of moles of molecules etc. depending on our needs.
4
Remember that because moles can be converted to mass, we can also produce a mass interpretation of a chemical equation.

5
Consider the reaction
2Na + Cl
2
Molecular interpretation:
 2NaCl
2 atom Na + 1 molecule Cl
2
 2 Formula Units of NaCl
Mole interpretation:
2 mole Na + 1 mole Cl
2
 2 moles of NaCl
Mass Interpretation:
2 mol x 23.0g/mol + 1mol x 71.0 g/mol 2molx58.5 g/mol
46.0 g Na + 71.0 g Cl
2
 117.0 g
117.0 g = 117.0 g

6

Examples
Show conservation of mass by determining the masses of all the reactants and products in the following reactions.
2 KBr(s) + Cl
2
(g)  2KCl(s) + Br
2
(l)
238.0 g 71.0g 149.2 g 159.8 g
309 g = 309 g
Fe
2
O
3
(s)+ 3CO(g)  2Fe(s) + 3CO
2
(g)
159.6 g 84.0g 111.6 g 132.0g
243.6 g = 243.6 g

7



A mole ratio is ratio between the number of moles of any two substances in a balanced chemical equation.
In the reaction
2 Al(s) + 3Br
2
(l)  2AlBr
3
(s)
What mole ratios can we write?
2 mol Al and 2 mol Al and 3 mol Br
2
3 mol Br
2
3mol Br
2 mol Al
2
2 mol AlBr
3 and 2 mol AlBr
3
2 mol AlBr
2
3 and 2 mol AlBr
2 mol Al 3 mol Br
3

 These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical
 Example: How many moles of chlorine are needed to react with 5 moles of sodium (without any sodium left over)?
2Na + Cl
2
 2NaCl
8
5 moles Na 1 mol Cl
2
2 mol Na
= 2.5 moles Cl
2

9
 Example: If 0.575 mole of CO
2 combustion of propane, C
3
H
8 is produced by the
, how many moles of oxygen are consumed? The balanced equation is:
? 0.575


C
3
H
8
+ 5 O
2
 3 CO
2
+ 4 H
2
O
ANALYSIS: Relating two compounds usually requires a mole-to-mole ratio
 SOLUTION:
0 .
575 mole CO
2 x
5 mole O
2
3 mole CO
2

0 .
958 moles O
2

10
Mole-Mole

Conversion Practice
Write the balanced reaction for hydrogen gas reacting with oxygen gas to make water.
1.
2 H
2
+ O
2
 2 H
2
O
To make 4.5 moles of water, how many mol of each reactant are needed?
2.
If we had 3.2 moles of oxygen, how many mol of hydrogen gas would we need?
3.
4.
If we had 52.2 moles of hydrogen, how mol of oxygen gas are needed?
If we had made 26.4 mol of water, how many mol of oxygen were used?

11
 Get into groups of not more than 5.
 Write a procedure or procedures for mass/mole and mass/mass conversions. You can make one procedure for the different types or you can make write two procedures.
 Follow your procedure EXACTLY as written to demonstrate how your group’s procedure(s) will solve the problems on the next slide.

12
1. How many moles of hydrogen gas can you make if you react 2.50 g of Na with water?
2. How many grams of aluminum oxide would you need to produce 7.12x10
3 moles of aluminum metal?
3. If you want to make 62.8 g of Mg
3 grams each of MgCl
2 and Na
3
PO
4
(PO
4
)
2, how many would you need?
4. If we combust 95.6 g of propane (C
3
H
8
):
A) How many grams of oxygen gas do we need?
B) How many grams of CO
2 are produced?

13
1. If we start with 95.6 g of propane (C
3
H
8
):
A) How many grams of oxygen gas do we need?
B) How many grams of CO
2
C
3
H
8
+ 5O
2 are produced?
 3CO
2
+ 4H
2
O
2. If you want to make 62.8 g of Mg
3 many grams each of MgCl
2
(PO need?
and Na
3
PO
4
4
)
2, how would you

27
 Remember Avogadro’s Law stated that the volume of a gas at constant pressure and temperature is directly proportional to the number of moles of gas.
Volume α n ; hold P & T constant
V = k x n
 This means that in a balanced reaction, the coefficients also can represent the ratio of volumes of gas, since equal volumes of gases at the same T & P have the same number of particles.

28
 From our reaction from before:
C
3
H
8
+ 5O
2
Also means the 1 L of C
3 of CO
2 and 4 L of H
2
O.
H
8
 3CO
2
+ 4H
2
O reacts with 5 L of O
2 to make 3 L
Example: What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C
3
H
8
)? Assume same T & P.
4.00 L C
3
H
8 x
5 L O
2
1 L C
3
H
8
= 20. L O
2
(g)

29
A. Determine the volume of hydrogen gas needed to react completely with 5.00 L of oxygen gas to form water. (Assume both gases at same T & P.)
Report answer to 1 decimal place in Quizdom.
2H
2
(g) + O
2
(g)  2H
2
O(l)

30
B. What volume of oxygen gas is needed to completely combust 2.36 L of methane gas
(CH
4
)?
Hint: write a balanced reaction first.
Input final answer to 3 SF’s in Quizdom.

31
 Can involve volumes, masses or moles!
 mass to moles by molar mass (usual)
 Volume to moles of gas using the ideal gas law (PV =nRT) or 22.4 L/mol
 Relate mass of solids (or liquids) to volume of gas
 Remember to convert temperature to
Kelvin if given as Celsius.

32
Ammonia is synthesized by the reaction below. If 5.0 L of nitrogen reacts completely with hydrogen at a pressure of
3.00 atm and 298 K, how many grams of ammonia are formed?
1.
2.
N
2
(g) + 3H
2
(g)  2NH
3
(g)
Know: V(N
2
) = 5.0 L; P = 3.00 atm; T = 298 K
Ask: What volume of NH
3 conditions? made under those
5.0 L N
2 x
2 L NH
3
1 L N
2
= 10.0 L NH
3

33
3.
Now that you know volume, can calculate moles.
n =
PV
RT
=
3.00 atm (10.0 L)
0.08206 L∗ atm mol∗k
(298K)
= 1.23 mol
NH
3
4.
1.23 mol NH
3 x 17 g/mol = 20.8 g NH
3

34
 Ammonium nitrate is a common ingredient in chemical fertilizers. Use the reaction below to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100 L of dinitrogen monoxide gas at STP (273 K and 1.0 atm).
NH
4
NO
3
(s)  N
2
O(g) + 2H
2
O(g)
Helpful info: Molar mass of NH
4
NO
3
= 80 g/mol
“ N
2
O = 44 g/mol
Procedure: Vol gas  mol gas  mol NH
4
NO
3
 g NH
4
NO
3

35
NH
4
NO
3
(s)  N
2
O(g) + 2H
2
O(g)
Helpful info: Molar mass of NH
4
NO
3
= 80 g/mol
“ N
2
O = 44 g/mol
Procedure: Vol gas  mol gas  mol NH
4
NO
3
 g NH
4
NO
3
1.0 atm∗0.100L
n = PV/RT =
0.08208
L−atm mol−K
∗(273 K) n = 0.00446 mol N
2
O*
1 mol NH
4
NO
3
1 mol N
2
O
= 0.357 g NH
4
NO
3
∗ 80 g mol

36
 When baking soda (sodium bicarbonate) decomposes, it gives off
CO
2
(g). If we decompose 1.50 g of baking soda, what volume of
CO
2 gas (in mL) could we collect at 1.00 atm and 298 K? (The molar mass of sodium bicarbonate is 84.0 g/mol).
2NaHCO
3
(s)  Na
2
CO
3
(s) + CO
2
(g) + H
2
O(g)
1.50 g * 1 mol/84.0 g * 1 mol CO
2
/2 mol NaHCO
3
= 0.00893 mol CO
2
=
Determine answer and enter in Quizdom with 3 SF’s.
= 0.00893 mol*(0.08206 L-atm/mol-K)*298 K/1.00 atm
= 0.218 L or 218 mL.

37



Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant.
That reactant is said to be in excess (there is too much).
The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant .




Try all of the reactants.
Calculate how much of a product we can get from each of the reactants
The reactant that makes the least amount of product is the limiting reactant .
38
 Be sure to pick a product! You can’t compare to see which is greater and which is lower unless you compare the results for the same product!

39
Limiting Reactants
The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when a reaction goes to completion.
The moles of product are always determined by the starting moles of the limiting reactant.

Which is the limiting reactant?
Strategy:
 Use the relationships from the balanced chemical equation
 You take each reactant in turn and ask how much product would be obtained, if each were totally consumed.
40
 The reactant that gives the smaller amount of product is the limiting reactant.

41
Limiting Reactant Example 1
If 20 mol hydrogen is reacted with 20 mol of oxygen to form water, which is the limiting reagent?
2H
2
+ O
2
2H
2
O
20 mol H
2 x
2 mol H
2
O
2 mol H
2

20 mol H
2
O
20 mol O
2 x
2 mol H
2
O
1 mol O
2

40 mol H
2
O
Only 20 mol of
H
2
O can be formed before
H
2 runs out!!

42
Limiting Reactant Example 2
When 100.0 g mercury is reacted with 100.0 g bromine to form mercury (II) bromide, which is the limiting reagent?
Hg + Br
2
HgBr
2 moles Hg

100 g Hg
1 mole Hg x
200 .
6 g Hg

0 .
498 moles Hg moles Br
2

100 g Br
2
1 mole x
159 .
8 g
Br
2
Br
2

0 .
625 moles Br
2

43
Limiting Reactant Example 2
 Moles of HgBr
2 produced:
 From Hg:
 From Br
2
:
0 .
498 moles Hg x
1 mole HgBr
1 mole Hg
2

0 .
498 moles HgBr
2
0 .
625 moles Br
2 x
1 mole HgBr
1 mole Br
2
2

0 .
625 mole HgBr
2
Thus the limiting reagent is
How much HgBr
2 is made?

44
 Iron(III) nitrate reacts with sodium hydroxide to make iron(III) hydroxide. If 0.750 moles of Fe(NO
3
)
3 are added to 2.00 moles of NaOH, how many moles of
Fe(OH)
3 are made?
Fe(NO
3
)
3
(aq) + 3NaOH  Fe(OH)
3
(s) + 3NaNO
3
(aq)
Determine your result and enter into Quizdom with 3
3 3 3
1 SF’s.
3
)
3
2 .
00 mol NaOH x
1 mol Fe(OH)
3
3 mol NaOH x

0 .
667 mol Fe(OH)
3

 15.0 g of potassium reacts with 45.0 g of iodine.
Determine which reactant is the limiting reactant and calculate how many grams of product are made.
2K(s) + I
2
(s)

2KI(s)
15 g K x
1 mole
39 .
1 g
K
K x
2 moles KI
2 K
 moles
45
45 g I
2 x
1 mole
253 .
8 g
I
2
I
2 x
2 moles KI
1 mole I
2

0 .
355 moles KI

58 .
9 g KI

46
 5.00 g of Cu metal react with a solution containing
20.0 g of AgNO
3 to produce Ag(s). Determine which reactant is the limiting reactant and calculate how much silver metal is made in grams.
Cu(s) + 2AgNO
3
(aq)

2Ag(s) + Cu(NO
3
)
2
(aq)
Determine your result and enter into Quizdom with 3 SF’s.
product made
5.00 g Cu x
1 mol Cu 2 mol Ag 107.9 g Ag x x
63.5 g 1 mol Cu mol Ag
= 17.0 g Ag
20.0g AgNO x
3
1 mol AgNO
3
169.9 g
x
2 mol Ag(s) 107.9 g Ag x
2 mol AgNO
3 mol Ag
= 12.7 g Ag
Therefore, AgNO
3 is the limiting reactant.

47
 By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. (Note: amount can mean either mole or mass.)
 Can we find the amount of excess potassium from Practice Problem 1?

Finding Amount of Excess Example - moles


We know from Practice Problem 1 that NaOH is LR, so
Fe(NO
3
)
3 is in excess. How much is left over in moles?
Fe(NO
3
)
3
(aq) + 3NaOH  Fe(OH)
3
(s) + 3NaNO
3
(aq)
 2.00 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 x
1 𝑚𝑜𝑙 𝐹𝑒 𝑁𝑂
3 3
3 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= 0.667 mol Fe(NO
3
)
3
USED!
48
0.750 mol Fe(NO3)3 give – 0.667 mol used = 0.083 mol
Fe(NO
3
)
3 leftover

Finding Amount of Excess Example - mass


15.0 g of potassium reacts with 45.0 g of iodine.
2 K + I
2
 2 KI
We found that Iodine is the limiting reactant, and 58.9 g of potassium iodide are produced.
49
45.0 g I
2
1 mol I
2
253.8 g I
2
2 mol K 39.1 g K
1 mol I
2
1 mol K
= 13.9 g K
USED!
15.0 g K – 13.9 g K = 1.1 g K EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should always start with it!

50
1.
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN
AMOUNT of reactants.
2.
Convert ALL of the reactants to the SAME product (pick any product you choose.)
3.
The reactant that gave you the lowest amount of product (in either grams or moles) is the
LIMITING REACTANT.

51
4.
The other reactant(s) are in EXCESS.
5.
To find the amount of excess, subtract the amount used from the given amount.
6.
If you have to find more than one product, be sure to start with the limiting reactant.
You don’t have to determine which is the
LR over and over again!

52
Limiting Reactant/Excess Reactant
Practice
5.00 g of Cu metal react with a solution containing
20.0 g of AgNO
3 to produce Ag(s). We know AgNO the limiting reactant from Practice Problem3.
3
Determine how much copper is leftover in grams.
Cu(s) + 2AgNO
3
(aq)  2Ag(s) + Cu(NO
3
)
2
(aq) is
Enter answer in Quizdom to 3 SF’s.

53
Theoretical and Actual Yields – Ch. 11.4



Theoretical yield is the amount of product predicted by the stoichiometry.
Actual yield is the amount actually obtained in an experiment.
Percent yield is the actual yield divided by the theoretical yield. It is a measure of the efficiency of your reaction. (How much bang for the buck!)
Percent Yield = Actual Yield x 100
Theoretical Yield

54
 Example: 0.500g of Mg was burned in excess oxygen and 0.752g of MgO was obtained.


Reaction: 2Mg(s) + O
2
(g)  2MgO(s)
What is the theoretical yield?
 0.500 g Mg x
1 mol Mg
24.3 g Mg x
2 mol MgO x
2 mol Mg
40.3 g MgO
=0.829 g mol MgO
MgO
 What is the actual yield?
 What was the percentage yield?

55
Theoretical and Actual Yield Practice 1
If we burn 8.00 g of CH
4
, what is the theoretical yield of water? If we only produce 17.0 grams of water, what is the percent yield? (Enter in
Quizdom to 1 decimal place.)
CH
4
+ 2O
2
 CO
2
+ 2H
2
O

56
Theoretical and Actual Yield Practice 2
Ethanol (C
2
H
5
OH) is produced from the fermentation of sugar (C
12
H
22
O
11
) in the presence of enzymes.
Determine the theoretical and percent yields of ethanol if 684 g of sucrose ferments to form 302 g of ethanol.
C
12
H
22
O
11
(aq) + H
2
O(l)  4C
2
H
5
OH(l) + 4CO
2
(g)