CHEMISTRY 104
CHAPTER 5: CHEMICAL REACTIONS
Homework problems: 17, 21, 23, 27, 31, 33, 35, 41, 49, 51, 53, 57, 61, 63, 67, 73, 75,
81, 83, 85, 89, 93
Outline
I.
Terms
A. Reactions
1. reactant(s)  product(s)
B. Molecular mass revisited
1. also called molecular weight, formula mass, formula weight
2. same concept as atomic mass, but for molecules instead of atoms
C. Mole
1. Avagadro’s number = 6.02214199… x 1023
2. just a number, like a dozen or a gross or a quarted
3. a very large number, but exactly what you need to go from amu
to grams
1.5 moles of CCl4
how many molecules of CCl4 in 1.5 moles of CCl4?
1.5 moles x (6.02 x 1023 molecules/mole) = 9.0 x 1023 molecules
How many atoms of C in 1.5 moles of CCl4?
1.5 moles x (6.02 x 1023 molecules/mole) x 1 C/ CCl4 = 9.0 x 1023 atoms C
How many of Cl in 1.5 moles of CCl4?
1.5 moles x (6.02 x 1023 molecules/mole) x 4 Cl/ CCl4 = 3.6 x 1024 atoms C
How many grams of CCl4 in 1.5 moles of CCl4?
1.5 moles x 153.8 g/mole = 230g
How many molecules of oxygen in 14.6 g of O2?
14.6 g x (1 mole/32.0g) x (6.02 x 1023 molecules/mole) =
2.75 x 1023 molecules
II.
Chemical Equations
A. Reactants and Products
B. Balanced by atoms AND charge AND mass
1. Coefficients
2. implied “1” if nothing written
a. like you to write it anyway for now
3. lowest whole number ratio
C. How to balance
1. method on p137 or…
2. another way
a. find biggest, ugliest molecule
b. put a “1” down as its coefficient
c. work your way back and forth from reactant to
product to get ratio with ugliest molecule
d. when all done, multiply through (if necessary) to
get whole number ratio
4. Cannot change formulas in eqn to balance it!
Start by picking a molecule. Put a one down as its coefficient. Then work back and
forth, balancing each element in turn. If you end up with a fractional coefficient,
multiply the whole thing through to get a whole number. It doesn’t matter which
molecule you pick to put a one as the coefficient. Usually, you pick the most
complicated molecule to make things easier. But it doesn’t matter which one you start
with, you will get the same answer.
Examples:
Balance the reaction below
Mg
+
HCl 
MgCl2
H2
Start by putting a one in front of hydrogen
Mg
+
HCl 
MgCl2
1 H2
one hydrogen molecule on the product side means two hydrogen atoms on the product
side, so we need two hydrogen atoms on the reactant side
Mg
+
2 HCl 
MgCl2
1 H2
Two chlorides on the reactant side means we need two chlorides on the product side, so
one magnesium chloride.
Mg
+
2 HCl 
1 MgCl2
1 H2
Finally, one Mg on the product side means one Mg on the reactant side
1 Mg
+
2 HCl 
1 MgCl2
1 H2
But what if I hadn’t picked hydrogen to put a “1” down as the coefficient? Would I get
the same answer? Let us try it.
The biggest, ugliest molecule is the magnesium chloride. Start by putting a one there.
Mg
+
HCl 
1 MgCl2
H2
One magnesium chloride means one Mg on the product side, so one Mg on reactant side
1 Mg
+
HCl 
1 MgCl2
H2
Now, we have gone down a dead end, so we need to go back to an earlier molecule that
already has a coefficient in front of it. You CANNOT put another random “1” down now
to continue. You HAVE to go back to something that is already got a coefficient. One
magnesium chloride also means two chlorides on the product side, so we need two on the
reactant side
1 Mg
+
2 HCl 
1 MgCl2
H2
Finally, two hydrogen atoms on the reactant side means 2 hydrogen atoms on the product
side. Two hydrogen atoms make one hydrogen molecule
1 Mg
+
2 HCl 
1 MgCl2
1 H2
Okay, but what if I accidentally pick the “wrong” molecule to start and put a one in front
of it? Let us try that. Start with a one in front of hydrogen monochloride.
Mg
+
1 HCl 
MgCl2
H2
One HCl means we have one hydrogen atom on the reactant side. That means we can
only have one hydrogen atom on the product side. So, how many hydrogen molecules
can we make with one hydrogen atom? We can only make half a hydrogen molecule.
Take the number of hydrogen atoms (1) and divide by the number of hydrogen atoms in
the molecule (2) to get one half (1/2). Just like when you go to the store and apples are
two dollars per pound, and you only have one dollar, so you can only get half a pound of
apples. It isn’t a whole number, but we can fix that later. For now, it is the correct ratio
of HCl to H2.
Mg
+
1 HCl 
MgCl2
½ H2
Now, were at another cul-de-sac, so we have to go back to another molecule that already
has a coefficient in front of it. We absolutely CANNOT pick another molecule and put a
one in front of that. If you do that, the equation will probably not balance. So, one HCl
means one chloride. One chloride atom means half a magnesium chloride
Mg
+
1 HCl 
½ MgCl2
½ H2
Finally, half a magnesium chloride means half a Mg on the product side, so we need half
a Mg on the reactant side.
½ Mg
+
1 HCl 
½ MgCl2
½ H2
Okay, so now we have the ratio, but it is not in whole numbers. Multiply both sides of
the equation by two.
1 Mg
+
2 HCl 
1 MgCl2
1 H2
Regardless of where you start, you will get the same answer. But you do need to start
somewhere.
Balance these equations:
Na
Al
CH4
C4H10
H3PO4
+
+
+
+
+
2 Na
4 Al
1 CH4
2 C4H10
1 H3PO4
H2O
O2
O2
O2
NaOH





+
+
+
+
+
2 H2O 
3 O2 
2 O2 
13 O2 
3 NaOH
NaOH +
Al2O3
H2O +
H2O +
Na3PO4
H2
CO2
CO2
+
H2O
2 NaOH
+
1 H2
2 Al2O3
2 H2O +
1 CO2
10 H2O
+
8 CO2

1 Na3PO4
+
3 H2O