Balancing Chemical Equations Academic Success Center Science Tutoring Area *

Balancing Chemical Equations
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Law of Conservation of Mass
“Matter cannot be created nor destroyed”
●Therefore the number of each type of atom on
each side of a chemical equation must be the same
●Balancing chemical equations is the process of
ensuring the conservation of matter
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Steps for Balancing an Equation
1.Write the correct formula for each of the reactants and
products
2.Verify the net ionic charge of each of the reactants and
products is balanced. If it is not balance it using subscripts
Example: Na+1Cl-1 + Ca+2SO4-2
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=
Na+1SO4-2 +
Ca+2Cl2-1
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Steps for Balancing an Equation
Note: Subscripts are only used to balance
charges within a molecule not to balance the number
of atoms on each side of an equation
3.Balance the number of atoms on each side of the equation
by changing the coefficient in front of the different reactants
and products
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Methods of Balancing Equations
Inspection Method or “Hit & Trial” Method
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List all the elements in the reaction (it is often
helpful to list them in a table of two columns - left side
column for reactants and right side column for
products
●Count the number of elements on each side
●Add coefficients to balance
●Also balance the ionic charges on each side
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Methods of Balancing
Equations
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Inspection Method Example
1
● K O + H O = KOH
2
2
● Each element has a
count of 2 on the
Reactant side (note one
oxygen in each reactant)
● To balance add a 2 in
front of the product
KOH
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R
P
K
2
1
O
2
1
H
2
1
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Methods of Balancing
Equations
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Inspection Method Example 2
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C3H8 + O2 = CO2 + H2O
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Note the odd numbers on
each side of the equation
and remember the hint to
multiply an odd number by
2 and then proceed
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To balance add a 3 to CO2,
a 4 to H2O and a 5 in front
of O2 on the reactant side
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R
P
C
3
1
H
8
2
O
2
3
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Helpful Hints
If an element appears in more than 1 place on one side of
the equation balance it last
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Ex. C6H12O6(s) + O2(g) = CO2(g) + H2O(l)
In this example oxygen is in two places on the products side of the equation
so leave oxygen for last
●Start with carbon, then hydrogen and finish up with oxygen
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Helpful Hints
If a fraction will balance the equation, multiply the entire
equation (both reactants and products) by the inverse of the
fraction to eliminate the fraction
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C3H7S(l) + O2(g) = CO2 + H2O(l) + SO2
●Placing 7/2 in front of water will balance the hydrogen on the
product side but it is best to multiple the entire equation through by 2 to
remove the denominator of the fraction to have the final balanced equation
with whole number coefficients
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Helpful Hints
If all coefficients are divisible by a small
whole number, divide to get the simplest
equation
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4C + 2O2 = 4CO simplified is 2C + O2 = 2CO
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Helpful Hints continued…
If polyatomic ions are the same on the reactant side as the product side,
balance the polyatomic ion as one group or one ion
●2(NH4)3PO4(aq) + 3 CaCl2 Ca3(PO4)2(s) + 6 NH4Cl(aq) 24 hydrogens, 6
nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines 24 hydrogens, 6
nitrogens, 2 phosphoruses, 8 oxygens, 3 calciums, 6 chlorines 6 ammoniums, 2
phosphates, 3 calciums, 6 chlorides 6 ammoniums, 2 phosphates, 3 calciums, 6
chlorides If an odd number of atoms appear on one side of the equation multiply the odd
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number by 2 and continue to balance the equation
Balance monatomic or diatomic elements last
●Balance H O last if it is in the equation
2
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Examples
Calcium carbonate + sulfuric acid yields calcium sulfate +
carbon dioxide + water
●CaCO3 + H2SO4 = CaSO4 + CO2 + H20
●Sodium chloride + ammonia gas + water + carbon dioxide
produces ammonium chloride + sodium bicarbonate
●NaCl + NH3 + H2O + CO2 = (NH4)Cl + NaHCO3
●Magnesium + water yields magnesium hydroxide +
hydrogen gas
●Mg + H2O = MgOH + H2
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