Chemical Reactions and Quantities
Chapter 6
The Chemical Equation
Reactants
→Products
The law of conservation of mass says that there must be the
same number of atoms of each element on both sides of the
equation.
MEMORIZE Table 2 pg. 209 (symbols in chemical equations)

NH
Cr
O

 4 2 2 7(s)   Cr2O3(s)  4 H 2O(l ) + N 2(g)
A chemical equation describes a
chemical change.
Any of the following indicate that a
chemical change has occurred:
Color change
Formation of a precipitate
Gas evolution
Heat absorbed or released
Light emission
To balance a chemical equation
adjust the coefficients.
When the following equation is balanced, what is
the sum of the coefficients?
Fe(NO 3 )3(aq)  K 2S(aq)  KNO 3(aq)  Fe 2S3(s)
The coefficients in a balanced equation must be
the smallest whole numbers that balance it.
A chemical equation can be
classified as one of the following:
Combination
Decomposition
Single replacement
Double replacement
Combustion
In a combination reaction 2 or more
elements or compounds combine to
form one product.
N2(g) + 3H2(g) → 2 NH3(g)
MgO(s) + CO2(g) → MgCO3(s)
Na2O(s) + H2O(l) → 2 NaOH(aq)
In a decomposition reaction a
reactant breaks apart into two or
more products.
CaCO3(s) → CaO(s) + CO2(g)
CaCl2·2H2O(s) → CaCl2(s) + 2 H2O(l)
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
In a single replacement reaction an
element (a metal or halogen) replaces
a like element in an ionic compound.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Br2(l ) + 2 KI(aq) → 2 KBr(s) + I2(s)
2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)
In a double replacement reaction two
ionic compounds react. The cations
switch places.
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq)
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l )
In a combustion reaction a carbon
containing compound reacts with
oxygen to form carbon dioxide and
water. Energy is always released.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
The mole (mol) is like a dozen (but
bigger)
1 mole _____ = 6.022 x 1023 _________
THIS IS A CONVERSION
FACTOR!
6.022 x 1023 is called Avogadro's number.
The molar mass of a substance is the
mass, in grams, of 1 mole of that
substance.
The molar mass of any element is the same number
as it's average atomic mass, but with units of g/mol.
So the molar mass of carbon is 12.011 g/mol.
This is a conversion factor!
1 mol carbon = 12.011 g carbon
The molar mass of a compound is
the sum of the molar masses of the
elements in it.
What is the molar mass of the sulfate ion?
We now have two conversion
factors we use a lot.
How many ants are there in 2.17 mol of ants?
How many grams of neon are there in 2.17 moles of
neon?
How many atoms of chromium are there in 3.142 grams
of chromium?
We can use these conversion
factors to answer some pre-lab
questions from experiment 10.
How many moles in 2.3 g of ethanol (C2H6O)?
How many moles in 0.476 g of NiCl2·6H2O?
What is the mass in grams of 4.5 x 10-3 moles of
CrCl3·6H2O?
Suppose you weigh out 4.023 g of an unknown hydrate.
You then perform some tests and find out that the anhydrous
salt is cobalt (II) nitrate.
Next, you heat up the 4.023 g of the hydrate until its
mass does not change anymore.
The remaining anhydrous salt has a mass of 2.529
grams. What is the formula of the unknown hydrate?
Stoichiometry
The coefficients in a BALANCED chemical equation give
us more conversion factors!
2 Al(s )  3 ZnSO 4(aq)  3 Zn (s )  Al2 (SO 4 )3(aq)
2 Al react with 3 ZnSO4 to produce
3 Zn and 1 Al2(SO4)3.
More stoichiometry.
2 Al(s )  3 ZnSO 4(aq)  3 Zn (s )  Al2 (SO 4 )3(aq)
2 moles Al = 3 moles ZnSO4
2 moles Al = 3 moles Zn
2 moles Al = 1 mole Al2(SO4)3
3 moles ZnSO4 = 3 moles Zn
3 moles ZnSO4 = 1 mole Al2(SO4)3
3 moles Zn = 1 mole Al2(SO4)3
2 Al(s )  3 ZnSO 4(aq)  3 Zn (s )  Al2 (SO 4 )3(aq)
Given this balanced equation, how many
moles of aluminum are required to react with
0.217 mol of zinc sulfate?
0.145 mol Al
We usually measure mass, so it is
more realistic to ask something like
this:
How many grams of aluminum sulfate can be
produced by the reaction of 5.199 grams of
zinc sulfate with excess aluminum?
2 Al(s )  3 ZnSO 4(aq)  3 Zn (s )  Al2 (SO 4 )3(aq)
The reactant that runs out first is
called the limiting reactant.
In a chemical reaction the same
principle applies:
To find the limiting reactant, do this:
Find moles of each reactant (do whatever it takes).
Divide moles of each reactant by its coefficient from the
balanced chemical equation.
The reactant that gives the smallest number from this
division is the limiting reactant.
What is the L.R. when 0.445 moles of
aluminum react with 0.601 moles of zinc
sulfate?
2 Al(s) + 3 ZnSO4(aq) → Al2(SO4)3(aq) + 3 Zn(s)
ZnSO4
What is the L.R. when 0.0536 grams of
aluminum react with 0.601 grams of zinc
sulfate?
2 Al(s) + 3 ZnSO4(aq) → Al2(SO4)3(aq) + 3 Zn(s)
Al
The moles of the L.R. is always the
starting point for all other dimensional
analysis involving that reaction!
Most problems involving limiting
reactants are dimensional
analysis!
How many grams of aluminum sulfate can
theoretically be produced by the reaction of
0.445 grams of aluminum with 0.601 grams of
zinc sulfate?
2 Al(s) + 3 ZnSO4(aq) → Al2(SO4)3(aq) + 3 Zn(s)
0.425 g Al2(SO4)3
How many grams of the excess reactant
are left once the reaction of 0.445 grams of
aluminum with 0.601 grams of zinc sulfate
is complete?
2 Al(s) + 3 ZnSO4(aq) → Al2(SO4)3(aq) + 3 Zn(s)
0.378 g Al
A chemical reaction almost never
goes perfectly!
actual yield
Percent yield=
×100
theoretical yield
What is the percent yield when 0.445 grams of
aluminum react with 0.601 grams of zinc sulfate
to produce 0.198 grams of aluminum sulfate?
2 Al(s) + 3 ZnSO4(aq) → Al2(SO4)3(aq) + 3 Zn(s)
46.6%