### Balancing Chemical EquationsProblems #1 - 10

Problem #1: FeCl3 + MgO ---> Fe2O3 + MgCl2

Solution:

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

2FeCl3 + MgO ---> Fe2O3 + 3MgCl2

The Fe also gets balanced in this step.

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.

Problem #2: Li + H3PO4 ---> H2 + Li3PO4

Solution:

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

Problem #4: FeS2 + Cl2 ---> FeCl3 + S2Cl2

Solution:

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

Problem #5: Fe + HC2H3O2 ---> Fe(C2H3O2)3 + H2

Solution:

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

Problem #6: H2(g) + V2O5(s) ---> V2O3(s) + H2O(ℓ)

Solution:

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.

Problem #7: HCl(aq) + MnO2(s) ---> MnCl2(aq) + Cl2(g) + H2O(ℓ)

Solution:

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.

Problem #8: Fe2O3(s) + C(s) ---> Fe(s) + CO2(g)

Solution:

1) Balance the iron:

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

2) Balance the oxygen:

Fe2O3(s) + C(s) ---> 2Fe(s) + 32CO2(g)

3) Balance the carbon:

Fe2O3(s) + 32C(s) ---> 2Fe(s) + 32CO2(g)

Note the 32 in front of the C and the CO2. What's that you say? You can't have 32 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 32 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 32 of a mole of carbon atoms or 32 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.

Problem #9: C5H11NH2 + O2 ---> CO2 + H2O + NO2

Solution:

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 132 in front of the water. I knew I'd have to eventually clear the 132, so I decided to do so right at the start.

2) Balance the nitrogen and the carbon:

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

3) Oxygen:

2C5H11NH2 + 372O2 ---> 10CO2 + 13H2O + 2NO2

4) Multiply through by 2 for:

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

Problem #10: CO2 + S8 ---> CS2 + SO2

Solution:

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

CO2 + 38S8 ---> CS2 + SO2

Most of the time the fraction used to balance is something with a 2 in the denominator: 12 or 52 or 132, for example. Not too often does one see 38. Pretty tricky!

2) Multiply through by 8:

8CO2 + 3S8 ---> 8CS2 + 8SO2

Bonus Problem: P4 + O2 ---> P2O3

Solution:

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):

P4 + O2 ---> 2P2O3

Then, the oxygen gets balanced:

P4 + 3O2 ---> 2P2O3

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

12P4 + O2 ---> P2O3

Let us balance the oxygen with a fraction as well:

12P4 + 32O2 ---> P2O3

Finally, multiply through by two:

P4 + 3O2 ---> 2P2O3

Balancing equations is fun!!

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