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Balancing Chemical Equations
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Background Information
Scientists use chemical equations as a shorthand way to represent the reactions that occur between substances.
Because of the law of conservation of matter, these equations must be balanced. In other words, the number of
atoms of each element must be conserved. For example, look at the reaction of hydrogen and oxygen to produce
water: H2 + O2 → H2O. The reactant side of the equation contains two atoms of hydrogen, as well as two atoms of
oxygen. On the product side, however, the hydrogen is conserved, but there is only one atom of oxygen. This
reaction is NOT balanced, and coefficients must be added to the equation. Coefficients are numbers that are put in
front of elements or compounds, within a chemical equation, to ensure that atoms are conserved. Since there are
2 atoms of oxygen on the left side of the equation, a coefficient of 2 is put in front of the water molecule: H 2 + O2
→ 2 H2O. But there's still a problem. The coefficient applies to all of the elements within the compound, so the
addition of the coefficient 2 doubles the number of hydrogen atoms as well. To remedy this, a 2 must also be
placed in front of the hydrogen on the left side of the equation. The balanced equation is, therefore, 2 H2 + O2 → 2
H2O. Always remember that the only way to balance a chemical equation is through the addition of coefficients.
You can NOT change the subscripts on an element or compound. Why?
ANSWER:
Balancing Method
Unfortunately, balancing equations isn't always as easy as the example given above. Sometimes, trying to balance
a chemical equation can be pretty difficult, particularly if you're just using a trial-and-error method.
Since chemists tend to be pretty logical, they ought to have come up with a more systematic way of balancing
chemical equations. Well, they have. Let's see how this method can be used to balance a chemical equation.
Suppose you were asked to balance the following equation:
C4H10 + O2 → CO2 + H2O
Using trial and error, you might get the right answer after several tries, or you might not. Instead, use the following
procedure to balance the equation:
1.
Rewrite the equation, with a blank line in front of each element or compound. The lines serve as
placeholders for the coefficients.
_____C4H10 + _____O2 → _____CO2 + _____H2O
2.
Represent each coefficient with a letter.
_____C4H10 +
_____O2 →
_____CO2 +
_____H2O
A
B
C
D
How many atoms of carbon are represented on the reactant side of the equation? Well, if you knew the value
of the coefficient for butane (C4H10), you would simply multiply it, by the subscript on carbon, to determine
the number of reactant carbon atoms. Since we don't know the value of butane's coefficient, the number of
atoms of carbon will have to be represented by the product of 4 and A. Thus, there are 4A carbon atoms on
the reactant side of the equation.
On the product side of the equation, there are 1C atoms of carbon (subscript x coefficient). And since matter
must be conserved in chemical reactions, the carbon atoms in this reaction must be conserved, so 4A must be
equal to 1C.
For each element, write an equation that represents the equality between the numbers of reactant and
product atoms.
Carbon: 4A = 1C
Hydrogen: 10A = 2D
Oxygen: 2B = 2C + 1D (Oxygen is found in both products, so the quantities
must be summed.)
These equations can then be solved for any of the variables, using simple arithmetic and substitution. Let's
express each of the variables in terms of A.
Carbon: 4A = 1C; this means that C = 4A
Hydrogen: 10A = 2D; dividing both sides by 2 shows that D = 5A
Oxygen: 2B = 2C + 1D; to solve for B, substitute the values for C and D
from above
2B = 2(4A) + 1(5A); now solve for B
2B = 8A + 5A
2B = 13A
B = 13/2 A
3.
Replace each of the non-A coefficients (B, C, and D) with its value, expressed in terms of A.
_____C4H10 +
_____O2 →
_____CO2 +
_____H2O
A
13/2A
4A
5A
Now, each of the coefficients is some multiple of A. What good does that do? Remember that coefficients
indicate the necessary ratios between substances in a chemical reaction. So the number of CO 2 molecules
produced is four times the number of butane molecules that react. We can get rid of the As by assuming that
A = 1.
4.
Replace each of the As with the number 1.
_____C4H10 +
_____O2 →
_____CO2 +
_____H2O
1
13/2
4
5
The resulting numbers represent the simplest ratios between the substances. When these ratios are not
expressed as whole numbers, each coefficient must then be multiplied by an integer that will remove any
and all fractions. All coefficients must be multiplied by the same integer; otherwise, the ratios between
substances will be altered.
5.
If necessary, multiply all coefficients by the same number, so that each is expressed as a whole number.
In this case, each coefficient must be multiplied by 2.
_____C4H10 +
_____O2 →
_____CO2 +
_____H2O
2
13
8
10
The resulting numbers are the coefficients for the balanced reaction. Always double-check your work by taking
an inventory of the elements to ensure that the reactant and product amounts are equivalent.
2 C4H10 +
13 O2 →
8 CO2 +
10 H2O
Each side of the reaction has 8 atoms of carbon, 20 atoms of hydrogen, and 26 atoms of oxygen.
The equation is balanced!
Exercises
Use the method outlined above to balance the following chemical equations. Show your work on a separate sheet
of paper.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
N2 + O2 + H2O → HNO3
NH3 + O2 → NO + H2O
C10H8 + O2 → CO2 + H2O
KO2 + CO2 → K2CO3 + O2
C6H12O6 + O2 → CO2 + H2O
FeS2 + O2 → Fe2O3 + SO2
NH3 + NO → N2 + H2O
C8H18 + O2 → CO2 + H2O
NiO + ClF3 → NiF2 + Cl2 + O2
C3H5(NO3)3 → N2 + O2 + CO2 + H2O
TiO2 + BrF3 → TiF4 + Br2 + O2
N2O4 + C9H20 → CO2 + H2O + N2
N2H3(CH3) + N2O4 → H2O + N2 + CO2
C7H5N3O6 → CO + H2 + N2 + C
C6H12O6 + KClO3 → CO2 + H2O + KCl
Ca3(PO4)2 + C + SiO2 → P4 + CaSiO3 + CO
NaCN + Au + O2 + H2O → NaAu(CN)2 + NaOH
Ba(OH)2 • 8H2O + NH4SCN → Ba(SCN)2 + H2O + NH3
NH4ClO4 + Al → Al2O3 + AlCl3 + NO + H2O
LiAlH4 + C(NO2)4 → Al2O3 + Li2O + CO2 + N2 + H2
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