Download PowerPoint - Balancing Equations

Chemistry
2/10/14
“Mystery creates wonder and wonder is the basis of man’s desire to
understand.” –Neil Armstrong
Do Now:
1. How many chorine atoms are in the following compound?:
4BaCl2
Agenda:
1. Students will be able to balance a chemical equation unassisted by
taking notes & working in groups
Announcements:
Science club is meeting tomorrow after school!
Introduction
Chemical reactions occur when bonds
between the outermost parts of atoms are
formed or broken
Chemical Equations
Their Job: Depict the kind of
reactants and products and their
relative amounts in a reaction.
4 Al (s) + 3 O2 (g) ---> 2 Al2O3 (s)
The numbers in the front are called
Coefficients
Chemical Equations
Because of the principle of the
conservation of matter,
an equation
must be
balanced.
It must have the same
number of atoms of the
same kind on both sides.
Lavoisier, 1788
Symbols Used in Equations
• Solid (s)
• Liquid (l)
• Gas (g)
• Aqueous solution (aq)
• + sign (combining)
• → ( yields or makes )
• Reactants - left of arrow
• Products- right of arrow
How molecules are symbolized
Cl2
2Cl
2Cl2
• Molecules may also have parenthesis to
indicate numbers of atoms. E.g. Ca(OH)2
• Notice that the OH is a group
O Ca O
H
• The 2 refers to both H and O H
• How many of each atom are in the following?
a) NaOH
Na = 1, O = 1, H = 1
b) Ca(OH)2 Ca = 1, O = 2, H = 2
c) 3Ca(OH)2 Ca = 3, O = 6, H = 6
Balancing Equations
– When balancing a chemical reaction you
may add coefficients in front of the
compounds to balance the reaction, but
you may
not
change the subscripts.
• Changing the subscripts changes the
compound. Subscripts are determined
by the valence electrons (charges for
ionic or sharing for covalent)
Subscripts vs. Coefficients
• The subscripts
tell you how
many atoms of
a particular
element are in a
compound. The
coefficient tells
you about the
quantity, or
number, of
molecules of
the compound.
Balancing equations: MgO
• The law of conservation of mass states that
matter can neither be created or destroyed
• Thus, atoms are neither created or destroyed,
only rearranged in a chemical reaction
• Thus, the number of a particular atom is the
same on both sides of a chemical equation
• Example: Magnesium + Oxygen
• Mg + O2  MgO Mg + O O  Mg O
• However, this is not balanced
• Left:
Mg = 1, O = 2
• Right: Mg = 1, O = 1
Balance equations by “inspection”
From
Mg + O2  MgO
2Mg + O2  2MgO
Mg + ½O2  MgO
Mg2 + O2  2MgO
4Mg + 2 O2  4MgO
is correct
is incorrect
is incorrect
is incorrect
Hints: start with elements that occur in one compound on each
side. Treat polyatomic ions that repeat as if they were a single
entity.
a) P4 + 5O2  P4O10
b) 2Li + 2H2O  H2 + 2LiOH
c) 2Bi(NO3)3 + 3K2S  Bi2S3 + 6KNO3
d) C2H6 + 3.5O2  2CO2 + 3H2O
2 C2H6 + 7 O2  4 CO2 + 6 H2O
K
+
H2O 
H2 +
KOH
• Reactants
• Products
•K
•K
•H
•H
•O
•O
Steps to Balancing Equations
There are four basic steps to balancing a chemical equation.
1. Write the correct formula for the reactants and the
products. DO NOT TRY TO BALANCE IT YET! You
must write the correct formulas first. And most
importantly, once you write them correctly DO NOT
CHANGE THE FORMULAS!
2. Find the number of atoms for each element on the left
side. Compare those against the number of the atoms
of the same element on the right side.
3. Determine where to place coefficients in front of
formulas so that the left side has the same number of
atoms as the right side for EACH element in order to
balance the equation.
4. Check your answer to see if:
– The numbers of atoms on both sides of the
equation are now balanced.
– The coefficients are in the lowest possible whole
number ratios. (reduced)
Some Suggestions to Help You
Some Helpful Hints for balancing equations:
• Take one element at a time, working left
to right except for H and O. Save H for
next to last, and O until last.
• IF everything balances except for O, and
there is no way to balance O with a whole
number, double all the coefficients and
try again. (Because O is diatomic as an
element)
Balance these skeleton equations:
a)
b)
c)
d)
e)
f)
g)
h)
i)
Mg + HCl  MgCl2 + H2
Ca + N2  Ca3N2
NH4NO3  N2O + H2O
BiCl3 + H2S  Bi2S3 + HCl
C4H10 + O2  CO2 + H2O
O2 + C6H12O6  CO2 + H2O
NO2 + H2O  HNO3 + NO
Cr2(SO4)3+ NaOH  Cr(OH)3+ Na2SO4
Al4C3 + H2O  CH4 + Al(OH)3
Balance these skeleton equations:
a)
b)
c)
d)
e)
f)
g)
h)
i)
Mg + 2HCl  MgCl2 + H2
3Ca + N2  Ca3N2
NH4NO3  N2O + 2H2O
2BiCl3 + 3H2S  Bi2S3 + 6HCl
2C4H10 + 13O2  8CO2 + 10H2O
6O2 + C6H12O6  6CO2 + 6H2O
3NO2 + H2O  2HNO3 + NO
Cr2(SO4)3+ 6NaOH  2Cr(OH)3+ 3Na2SO4
Al4C3 + 12H2O  3CH4 + 4Al(OH)3
a)
b)
c)
d)
e)
f)
g)
h)
Li + H2O  H2 + LiOH
P4 + O2  P4O10
C2H6 + O2  CO2 + H2O
CS2 + O2  CO2 + SO2
AsCl3 + H2S  As2S3 + HCl
AgNO3 + FeCl3  AgCl + Fe(NO3)3
KClO3 KCl + O2
SO2 + O2  SO3
a)
b)
c)
d)
e)
f)
g)
h)
2Li + 2H2O  H2 + 2LiOH
P4 + 5O2  P4O10
2C2H6 + 7O2  4CO2 + 6H2O
CS2 + 3O2  CO2 + 2SO2
2AsCl3 + 3H2S  As2S3 + 6HCl
3AgNO3 + FeCl3  3AgCl + Fe(NO3)3
2KClO3  2KCl + 3O2
2SO2 + O2  2SO3