Balancing chemical equations difficult examples

1) Balance the Cl (note that 2 x 3 = 3 x 2):

2) Pick either the O or the Mg to balance next:

2FeCl₃ + 3MgO ---> Fe₂O₃ + 3MgCl₂

The other element (Mg or O, depending on which one you picked) also gets balanced in this step.

1) Balance the Li:

3Li + H₃PO₄ ---> H₂ + Li₃PO₄

2) Now, look at the hydrogens. See how the H comes only in groups of 3 on the left and only in groups of 2 on the right? Do this:

3Li + 2H₃PO₄ ---> 3H₂ + Li₃PO₄

Remember 2 x 3 = 6 and 3 x 2 = 6. It shows up a lot in balancing problems (if you haven't already figured that out!).

3) Balance the phosphate as a group:

3Li + 2H₃PO₄ ---> 3H₂ + 2Li₃PO₄

4) Oops, that messed up the lithium, so we fix it:

6Li + 2H₃PO₄ ---> 3H₂ + 2Li₃PO₄

Problem #3: ZnS + O₂ ---> ZnO + SO₂

1) Balance the oxygen with a fractional coefficient (Zn and S are already balanced):

ZnS + (3/2)O₂ ---> ZnO + SO₂

2) Multiply through to clear the fraction:

2ZnS + 3O₂ ---> 2ZnO + 2SO₂

1) See how the Fe and the S are already balanced? So, look just at the Cl. There are a total of 5 on the right-hand side, so we put 5 on the left:

FeS₂ + (5/2)Cl₂ ---> FeCl₃ + S₂Cl₂

2) Clear the fraction by multiplying through by 2:

2FeS₂ + 5Cl₂ ---> 2FeCl₃ + 2S₂Cl₂

1) Balance the acetate:

Fe + 3HC₂H₃O₂ ---> Fe(C₂H₃O₂)₃ + H₂

2) Balance the hydrogen:

Fe + 3HC₂H₃O₂ ---> Fe(C₂H₃O₂)₃ + (3/2)H₂

3) Clear the fraction:

2Fe + 6HC₂H₃O₂ ---> 2Fe(C₂H₃O₂)₃ + 3H₂

1) Balance the oxygen:

H₂(g) + V₂O₅(s) ---> V₂O₃(s) + 2H₂O(ℓ)

2) Balance the hydrogen:

2H₂(g) + V₂O₅(s) ---> V₂O₃(s) + 2H₂O(ℓ)

Note that the vanadium was not addressed because it stayed in balance the entire time. Note how the hydrogen started out balanced, but the balancing of oxygen affected the hydrogen, which we addressed in the second step.

1) Balance the chlorine:

4HCl(aq) + MnO₂(s) ---> MnCl₂(aq) + Cl₂(g) + H₂O(ℓ)

2) Balance the hydrogen:

4HCl(aq) + MnO₂(s) ---> MnCl₂(aq) + Cl₂(g) + 2H₂O(ℓ)

With this last step, the oxygen is also balanced and the Mn was never mentioned because it started out balanced and stayed that way.

1) Balance the iron:

Fe₂O₃(s) + C(s) ---> 2Fe(s) + CO₂(g)

2) Balance the oxygen:

Fe₂O₃(s) + C(s) ---> 2Fe(s) + 3 ⁄₂ CO₂(g)

3) Balance the carbon:

Fe₂O₃(s) + 3 ⁄₂ C(s) ---> 2Fe(s) + 3 ⁄₂ CO₂(g)

Note the 3 ⁄₂ in front of the C and the CO₂. What's that you say? You can't have 3 ⁄₂ of an atom? Ah, just you wait.

4) Multiply through by two for the final answer:

2Fe₂O₃(s) + 3C(s) ---> 4Fe(s) + 3CO₂(g)

Comment: one way to look at this is that using the 3 ⁄₂ was just a mathematical artifice to balance the equation. The chemical reality of atoms reacting in ratios of small whole numbers is reflected in the final answer.

Another way to look at the coefficients is in terms of moles. We can certainly have 3 ⁄₂ of a mole of carbon atoms or 3 ⁄₂ of a mole of carbon dioxide molecules. The final step towards whole number coefficients is just a convention. The chemical equation is balanced in a chemically-correct sense with the fractional coefficients.

1) Balance the hydrogens first:

2C₅H₁₁NH₂ + O₂ ---> CO₂ + 13H₂O + NO₂

Notice that I used a 2 in front of C₅H₁₁NH₂. That's because I knew that there are 13 hydrogens in the C₅H₁₁NH₂ and that meant a 13 ⁄₂ in front of the water. I knew I'd have to eventually clear the 13 ⁄₂ , so I decided to do so right at the start.

2) Balance the nitrogen and the carbon:

2C₅H₁₁NH₂ + O₂ ---> 10CO₂ + 13H₂O + 2NO₂

2C₅H₁₁NH₂ + 37 ⁄₂ O₂ ---> 10CO₂ + 13H₂O + 2NO₂

4) Multiply through by 2 for:

4, 37 ---> 20, 26, 4

1) The only thing not balanced already is the S:

CO₂ + 3 ⁄₈ S₈ ---> CS₂ + SO₂

Most of the time the fraction used to balance is something with a 2 in the denominator: 1 ⁄₂ or 5 ⁄₂ or 13 ⁄₂ , for example. Not too often does one see 3 ⁄₈ . Pretty tricky!

2) Multiply through by 8:

8CO₂ + 3S₈ ---> 8CS₂ + 8SO₂

1) Suppose you decide to balance the oxygen first:

P₄ + 3O₂ ---> 2P₂O₃

This depends on seeing that the oxygen on the left comes in twos and the oxygen on the right comes in threes. So, you use a three and a two to arrive at six oxygens on each side. Least common multiple, baby!!

2) Suppose you balance the phosphorus first (with a whole number):

3) Suppose you balance the phosphorus first (with a fraction):

Balancing equations is fun!!