Chapter 6 Reaction Calculations
6-1 The brownie analogy
6-2 Proportional Relationships
2Mg
2
+
:
(1)O2
1
+
:
2MgO
2
20
10
20
23
23
2(6X10 ) atoms : 1(6x10 ) atoms : 2(6x1023) molecules
2 mol atoms
:
1 mol atoms
:
2 mol molecules
Mole Ratio
- used to predict what would happen if the amount of one of
these substances is changed.
Eg. In above
- How many moles of O2 would react with 1 mole of Mg to form
MgO?
Ratio = 2 mol Mg = 1 mol Mg
1 mol O2 x mol O2
x mol O2 = 1 mol O2 x 1 mol Mg
2 mol Mg
= 0.5
- How many moles of MgO could be produced from 1 mole Mg?
Ratio = 2 mol MgO = x mol MgO = 1 mol
2 mol Mg
1 mol Mg
1
Instant Practice –
How many moles of MgO would be produced from 0.175 moles O2?
Given: 2 Mg + 1 O2 Æ 2 MgO
1. Mole Ratio
2:1:2
2 mol MgO = x mol MgO
1 mol O2 0.175 mol O2
= 0.35 mol
Question: How many g of MgO?
0.35 mol x
40.3 g MgO
mol
= 14.11g
6-3 Grams Æ Moles Æ Atoms
How many molecules of MgO would be produced from the oxidation
of 2.00g of Mg?
Given: 2.00g Mg
Find: Molecules of MgO
1) Write balanced equation
2Mg + O2 Æ 2 MgO
2) Mole ratio
2:1:2
3) Find moles of Mg reacted (oxidized)
2.00g Mg x 1 mol = 0.0823 mol
24.31g Mg
4) Apply mole ratio to determine moles of MgO
2
2 mol MgO = x mol MgO
=
2 mol Mg
0.0823 mol MgO
0.0823 mol
5) Convert mol MgO to molecules
0.0823 mol x 6.02 x 1023 molecules U = 4.95 x 1022
mol
Exercises p.154 #2,3,4
6-4 Stoichio - metry
- elements
- measurements
- calculating or measuring elements or compounds involved in a
chemical change
Exercises p. 154 Review #1,2 on board; #3-6 instant practice
6-5 Mole - Mass calculations
Example 6.2
How many moles of sodium metal would be needed to react with
chlorine gas to make 735g of sodium chloride?
1) Write and balance equation
2 Na(s) + Cl2(g) Æ 2 NaCl(s)
2) mol ratio = 2:1:2
3) Find mol NaCl
737g x
1 mol
= 12.60 mol
58.5g NaCl
4) Using mol ratio determine mol Na
3
2 mol Na = x mol Na = 12.60 mol Na
2 mo NaCl 12.60mol NaCl
5) Determine mass Na
12.60 mol Na x 23.0 g Na = 289.8g
mol
sig figs - 290g
6-6 Mass - Mass Calculation
How many grams of potassium chloride, KCl, are produced by
decomposing 118g of potassium chlorate, KClO3?
1) Write balanced equation
KClO3 Æ KCl + 3O
2) Mol ratio
2: 1: 3
3) Determine mol of KClO3
118g KClO3 x 1 mol = 0.96 mol
122.6g KClO3
Side calc.
H - 1 x 39.1 = 39.1
Cl - 1 x 35.5 = 35.5
O - 3 x 16.0 = 48.0
122.6
2 KClO3 Æ 2 KCl + 3 O2
0.96 mol Æ ??
4) Using mol ratio - determine mol KCl
2 mol KCl = x mol KCl = 0.96 mol KClO3
2 mol KClO3 0.96 mol KClO3
5) Determine g KCl
4
0.96 mol KCl x
6-7
74.6 g
mol
= 71.6 g KCl
Molarity and Replacement
Review: molarity (M) = moles solute
1 Litre solution
so…
Given: 250mL of 3.0M AgNO3(aq) (or any other)
= 3.0 moles = x moles =
1 Litre
0.250 L
3.0 moles x 0.250L = 0.75 moles
???Okay???
Problem: How many grams of copper will react to completely replace
sliver from 0.100M solution of silver nitrate, AgNO3?
1) Write balanced equation
Cu + AgNO3 Æ Cu(NO3)2 + 2 Ag
2) Mol ratio
1:2:1:2
3) Determine mol AgNO3
0.100 mol AgNO3 = x mol AgNO3 = 0.021 mol
1L
0.208 L
4) Determine mol Cu (using mol ratio)
5
1 mol Cu
=
x mol Cu
2 mol AgNO3
0.0208 mol AgNO3
= 0.0104
5) Determine g Cu
0.0104 Mol Cu x
6-8
63.5 g
mol
= 0.660g
Limiting reagents (reactants) and reactants “in Xs”
- an excess reactants is one that is added in larger then
necessary amounts
- limiting reactants determines (or limits) the amount of products
Eg.
Zn +
Given:
1.21mol
2HCl
Æ
ZnCl2
+
H2
2.65mol
Find: Which reactant is in excess?
How much in excess?
How many g ZnCl2 produced?
How many g H2 produced?
1) What is the mole ratio?
1 Zn : 2 HCl
2) Determine how much HCl would be required to react with 1.21 mol
Zn (using mol ratio)?
6
2 mol HCl = x mol HCl = 2.42 mol HCl
1 mol Zn
1.21 mol Zn
3) Determine excess HCl
2.65 HCl - 2.42 mol HCl = 0.23 mol NXS
4) If HCl is NXS the Zn must be limiting reactant
5) Determine moles ZnCl2 using mol ratio
1 mol ZnCl2 = x mol Zn Cl2 = 1.21 mol ZnCl2
1 mol Zn
1.21 mol Zn
6) Determine moles H2 using mol ratio
1 mol H2 = x mol H2 = 1.21 mol H2
1 mol Zn 1.21 mol Zn
Example 6.9
When 79.1 g of Zinc are reacted with 1.05L of 2.00M
hydrochloric acid, HCl, to produce zinc chloride, ZnCl2, and
hydrogen gas, H2, which reactant will be in excess and by how
much? Calculate mass of each product.
1. Write balanced equation
Zn + 2 HCl Æ ZnCl2 + H2
2. What-to-what
79.1%g ( Zn )+ 1.05L of 2.00M ( 2 HCl )
Å
?excess?
Æ
7
3. Determine moles of Zn available
79.1 g Zn x
1 mol = 1.21 mol
65.4g Zn
4. Determine moles of HCl available
2.00 moles = x moles = 2.10 mol
1L
1.05L
5) Determine mol ratio
Zn + 2 HCl Æ ZnCl2 + H2
1
2
6. a) Determine moles of HCl needed to react with 1.21 mol Zn.
(using mol ratio)
2 mol HCl = x mol HCl = 2.42 mol HCl
1 mol Zn
1.21 Zn
* 2.42 mol HCl is needed to react with 1.21 mol Zn.
However, there is only 2.10 mol HCl available. HCl is
the limiting reactant.
6. b) Determine moles of Zn needed to react with 2.10 mol HCl.
1 mol Zn = x mol Zn = 1.05 mol
2 mol HCl 2.10 mol HCl
* There is 1.21 mol Zn available but we only need 1.05 mol Zn
to react with 2.10 mol HCl. Therefore, 1.21g - 1.05g Zn = 0.16
mol zinc is in XS.
8
7. Rewrite balanced equation noting mole ratio and mol of reactants
involved in actual Rxn.
Zn + 2HCl Æ ZnCl2 + H2
1 : 2
: 1
:1
8. a) Determine mol ZnCl2 formed using mol ratio
b) Determine g ZnCl2
1.05 mol x
136.4 g
mol
= 143.2g
9. a) Determine mol H2 formed using mol ratio
1.05 mol x 2.02g H2 = 2.12g
mol
Percent Yield
- Rarely…does a reaction work in the lab where you get the
amount of product that you think you should. Usually, it is far
less. This is expressed in Percent Yield.
- Theoretical Yield - what the balanced equation says you should get
- Actual Yield - what you really get
- Percent Yield -
__ actual __
theoretical
x100%
Example 6-10
9
When 45.8g of K2CO3 are reacted completely with excess HCl, 46.3g
KCl are produced. H2O and CO2 is also formed. Calculate theoretical
and percent yield of KCl.
1. Write balanced equation
K2CO3 + HCl Æ KCl + H2O + CO2
mol ratio 1
: 2
: 2
:
1
: 1
2. Determine mol K2CO3
45.8g K2CO3 x 1 mol = 0.33 mol
138.2g K2CO3
K
C
O
Side calc
2 x 39.1 = 78.2
1 x 12.0 = 12.0
3 x 16.0 = 48.0
138.2
3. Determine theoretical yield of KCl (using mol ratio)
2 mol KCl = x mol KCl = 0.66 mol
1 mol K2CO3 0.33 mol K2CO3
Æ 0.66 mol KCl x 74.6g KCl = 49.24g
mol KCl
4. Determine Percent Yield
Actual (given) x 100% = 46.3g x 100% = 94%
Theoretical (calc)
49.24g
10
How to Page Solving Dilution Problems
There are two ways to go about this:
C1v1=c2v2
Or
Rationalize your way through.
The following is dedicated to the latter method:
Steps
1) Find moles of solute required in final solution
2) Apply molar ratio: moles (start) = x moles (finish)
L (start)
xL
(finish)
3) Solve for X
Eg. Describe how you would prepare a 750mL quantity of 0.10M
NaIO3 from a 0.20M
1. Moles NaIO3 in final volume
0.10 moles x 0.750L = 0.075 moles
1L
(find M)
(find V) (find mol of solute req’d)
2. 0.20 moles = 0.075 moles
L
x Litres
(flip)
L
= x Litres = 1 L x 0.075 mol = 0.375 L
0.02 moles
0.075 moles
0.20 mol
or: 375 mL
To 375mL of stock solution, add enough water to make 750mL ( ie
11
750mL - 375mL = 375mL
Vf
- stock solution = water
The How to Sheet
Solving Mole Ratio Problems
1. Moles to Grams
Problem:
How many grams of MgO would be produced by the oxidation of
0.175 moles of O2 magnesium?
Given: 0.175 moles of O2 Æ grams MgO
1) Write and balance equation
2Mg + O2 Æ 2MgO
2) Determine mole ratio
3) Plot a road map
2Mg
+
O2
0.175 moles O2
(convert)
a)
Æ
Æ
2MgO
grams MgO
convert)
moles O2Æ mol ratio Æ moles MgO
a) mole ratio
2 mol MgO = x mol MgO
1 mol O2
0.175 mol O2
x = 0.35 mol MgO
12
b) convert
0.35 mol x 40.3 g Mg = 14.11g
mol
2. Grams to grams
Problem:
How many grams of MgO would be produced from the oxidation of
2.00g of Mg?
Given: 2.00g Mg Æ x g MgO
1) Write and balance equation
2Mg
+
O2
Æ
2MgO
O2
Æ
2MgO
Æ
g MgO
2) Determine mol ratio
3) Plot a road map
2Mg
+
2.00g
(convert)
Æ
moles Mg
moles MgO
a) Convert 2.00g Æ mol Mg
2.00g x
1 mol = 0.082 mol Mg
24.3g Mg
b) Mole ratio Mg Æ MgO
2 mol MgO = x mol MgO
x = 0.082 mol
13
2 mol Mg
0.082 mol Mg
c) Convert mol MgO Æ g MgO
0.082 mol MgO x 40.0g MgO = 3.29g MgO
mol
3. Grams to molecules
Problem: How many molecules of MgO would be produced from the
oxidation of 2.00g of Mg?
Given: 2.00g Mg Æ molecules MgO
1) Write and balance equation
2) Determine the mol ratio
3) Plot a road map
2.00g Mg
Æ
molecules MgO
a) convert
c) convert
mol Mg Æ b) mol ratio Æ
mol MgO
a) Convert 2.00g Æ mol Mg
2.00g x 1 mol = 0.082 mol Mg
24.3g Mg
b) Mol ratio Mg Æ MgO
2 mol MgO = x mol MgO
2 mol Mg
0.082 mol Mg
x = 0.082 mol
14
c) Convert mol MgO Æ molecules MgO
0.082 mol MgO x 6.02 x 1023 molecules
mol
= 4.94 x 1022 molecules
Mol - Mass*
*= pg. 164 (1-8) Use Road Map
Mass - Mass*
More How-To For Molarity and Replacement
Reactions
Example 6.6 The good old Silver Nitrate replacement reaction!
Given: 208mL of 0.100M AgNo3(aq)
Find: g Cu to completely replace AgNo3(aq)
1) Write and balance equation
2AgNo3(aq)
+
Cu(s) -->Cu(NO3)2 (aq) + 2Ag(s)
+
Cu(s) -->Cu(NO3)2 (aq) + 2Ag(s)
2) Road Map
2AgNo3(aq)
208mL 0.100M AgNo3(aq)
a)
Æ g Cu
c)
15
mol AgNo3(aq)
Æ b)
mol ratio Æ mol Cu
a) 208mL 0.100 M Æ mol AgNo3(aq)
0.208L x 0.100M = 0.0208 mol
L
b) 2 mol AgNo3(aq) Æ 1 mol Cu
2 mol AgNo3(aq) = 0.0208 mol AgNo3(aq)
1 mol Cu
x mol Cu
Æ Flip and multiply
1 mol Cu x 0.0208 mol AgNo3(aq) = x mol Cu = 0.0104 mol
2 mol AgNo3(aq)
c) mol Cu Æ g Cu
0.0104 mol x 63.5g = 0.660g
mol
Molarity
Flavor #1 - Given grams of a solute in a volume of solution - find
molarity
Problem: What is the molarity of a 125mL solution containing 25g of
silver nitrate ( AgNo3 )
16
(c)
(b)
Molarity = moles
Litre
Å 25g AgNo3
(a) molecular wt AgNo3 from PT
Given: 125mL = 0.125L
a) Molecular Wt. AgNo3
b) Moles AgNo3
Ag
N
O
= 25g AgNo3
169.9 g/mol AgNo3
1x107.89
1x14.01
3x16.00
169.9
= 0.147 mol
c) Molarity = 0.147 mol
0.125 L
= 1.18 M
Flavor #2 - Given the molarity of a solution and a volume - find
“corresponding” grams
17
Problem: How many grams of copper sulphate ( CuSO4 ) are in 75mL
of a 5 M solution?
1) Set up a molar ratio
(5 M CuSO4)
5 mol CuSO4 =
IL
x mol CuSO4
0.075L
x = 0.376 moles CuSO4 (in 75mL of CuSO4)
2) Convert moles to grams
molecular Wt. CuSO4
i) Cu 1x63.55 = 63.55
S 1x32.06 = 32.06
O 4x16.00 = 64.00
160.61
ii) grams = mol x molecular Wt.
= 0.375 x 160.61g/mol
= 60.23 g or 60 g (Sig Figs)
Flavor #3 - Given the molarity of a solution and the mass of a solute find corresponding volume
18
Problem: What volume of a 1.5M solution would you find 3 grams of
sodium chloride?
1) Set up molar ratio
1.5M NaCl
1.5 mol NaCl = moles NaCl
1L
xL
2) Convert grams to mol
Å(3g)
i) Na 1x22.99 = 22.99
Cl 1x35.45 = 35.45
58. 44
ii) moles =
grams NaCl
molecular wt. NaCl
=
0.05 mol
iii) 1.5 mol NaCl = 0.05 mol NaCl
1L
xL
* flip and solve for X
= 0.075 L (75mL)
END- How-to Sheet
19