CHEMICAL EQUATIONS
DOE-HDBK-1015/1-93
Fundamentals of Chemistry
CH-01
Rev. 0
Page 42
Example 1:
FeS + O Fe O + SO
2
2
2
3
2
Solution 1:
Starting with Fe O (see guideline b), write 2FeS + O Fe O + SO , which
2
3
2
2
2
3
2
balances the Fe on each side of the equation.
Now there are 4 S atoms on the left side so balance the S atoms by writing
2FeS + O Fe O + 4SO .
2
2
2
3
2
Everything is balanced except the O. There are now 2 O atoms on the left and 11 O
atoms on the right. To get 11 O atoms on the left write
2FeS + 5.5O Fe O + 4SO .
2
2
2
3
2
This makes both sides of the equation balanced except the coefficients must be whole
numbers (guideline f). To meet guideline f, multiply both sides by two which will bring
the left side to a whole number of O molecules.
2
Thus, the solution is 4FeS + 11O 2Fe O + 8SO .
2
2
2
3
2
Example 2:
NH + CuO H O + N + Cu
3
2
2
Solution 2:
Start with NH since there are two N atoms on the right of the equation. To balance the
3
N atoms write 2NH + CuO H O + N + Cu (guideline b).
3
2
2
Since the H appears in only the NH and H O and the NH has been balanced, the H O
3
2
3
2
will be balanced. Write 2NH + CuO 3H O + N + Cu (guideline e).
3
2
2
Oxygen appears only in CuO and in H O, and since the H O has been already been
2
2
balanced write 2NH + 3CuO 3H O + N + Cu (guideline e).
3
2
2
That leaves the Cu to be balanced. Thus, the solution is
2NH + 3CuO 3H O + N + 3Cu.
3
2
2
