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Examples of Balancing Chemical Equations
This is a set of worked examples of how to balance chemical equations, a very important skill in chemistry. Try each question first, using a pen and paper. Then click the word Answer and magically all will be revealed, without even leaving this page. Click Answer a second time to close the answer and move on to the next question. If the answer is long, scroll down until the example is at the top of the screen before clicking it. I hope you enjoy this chemistry tutorial.
C5H12 + O2 ---> CO2 + H2O
C5H12 + O2 ---> 5CO2 + H2O
There are twelve hydrogens on the left but only two on the right hand side, and hydrogen is in a single species on each side. Put a 6 in front of the H2O on the right hand side.
C5H12 + O2 ---> 5CO2 + 6H2O
Finally, there are only two oxygens on the left hand side but 16 of them on the right hand side. So put a 8 in front of the O2 on the left hand side.
C5H12 + 8O2 ---> 5CO2 + 6H2O
It's now a balanced chemical equation.
Zn + HCl ---> ZnCl2 + H2
Zn + 2HCl ---> ZnCl2 + H2
And if you look carefully, you will see that the equation is now balanced, with one Zn on each side, two hydrogens on each side and two chlorines on each side. Some examples can be rather easy!
Ca(OH)2 + H3PO4 ---> Ca3(PO4)2 + H2O
3Ca(OH)2 + H3PO4 ---> Ca3(PO4)2 + H2O
There are two PO4 ions on the right but only one on the left side, and the P doesn't appear anywhere else (so the group remains intact). Put a 2 in front of the H3PO4 on the left side.
3Ca(OH)2 + 2H3PO4 ---> Ca3(PO4)2 + H2O
Finally, there are six oxygens on the left hand side not present as PO4 but only one on the right hand side not in the PO4. So put a 6 in front of the H2O on the right hand side.
3Ca(OH)2 + 2H3PO4 ---> Ca3(PO4)2 + 6H2O
It's now a balanced equation. Note how we treated the PO4 ion as a single species to be balanced.
FeCl3 + NH4OH ---> Fe(OH)3 + NH4Cl
FeCl3 + NH4OH ---> Fe(OH)3 + 3NH4Cl
The next most obvious unbalanced part is that there are now three NH4 groups on the right but only one on the left hand side. So put a 3 in front of the NH4OH on the left.
FeCl3 + 3NH4OH ---> Fe(OH)3 + 3NH4Cl
And if you count up the atoms on each side, you will see that this is now a balanced chemical equation.
S8 + F2 ---> SF6
S8 + F2 ---> 8SF6
And now we can see that there are 48 fluorines on the right and only two on the left, so put a 24 in front of the F2 on the left.
S8 + 24F2 ---> 8SF6
And that chemical equation is now balanced. Check this by counting the number of atoms of each type on each side.
C2H6 + O2 ---> CO2 + H2O
C2H6 + O2 ---> 2CO2 + H2O
Because the oxygen is in two compounds on the right, we will look at the hydrogen next as it is in one compound on each side of the equation. There are six hydrogens on the left and two on the right - put a 3 in front of the H2O on the right
C2H6 + O2 ---> 2CO2 + 3H2O
Now we have two oxygens on the left and seven on the right. Put a 3 1/2 in front of the O2 on the left.
C2H6 + 3 1/2O2 ---> 2CO2 + 3H2O
BUT we don't like having a half in a chemical equation, so multiply every coefficient on both sides by two.
2C2H6 + 7O2 ---> 4CO2 + 6H2O
And this is now a balanced chemical equation.
Al2(CO3)3 + H3PO4 ---> AlPO4 + CO2 + H2O
Al2(CO3)3 + H3PO4 ---> 2AlPO4 + CO2 + H2O
There are now two PO4 units on the right, and only one on the left, and there is no other phosphorus containing species, so let's make it 2H3PO4 on the left
Al2(CO3)3 + 2H3PO4 ---> 2AlPO4 + CO2 + H2O
There are three carbons on the left, and only one on the right, so we need to make it 3CO2 on the right
Al2(CO3)3 + 2H3PO4 ---> 2AlPO4 + 3CO2 + H2O
Nearly done, but there are six hydrogens on the left and only two on the right, so it should be 3H2O on the right
Al2(CO3)3 + 2H3PO4 ---> 2AlPO4 + 3CO2 + 3H2O
And if you count them up carefully, there are now 17 oxygens on each side of the equation, so it is now balanced.
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