Problems #1 - 10

Fifteen examples | Problems 11-25 | Problems 26-45 | Problems 46-65 |

Six "balancing by groups" problems | Only the problems | Return to Equations Menu | |

Sixteen balance redox equations by sight |

**Problem #1:** FeCl_{3} + MgO ---> Fe_{2}O_{3} + MgCl_{2}

**Solution:**

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

2FeCl_{3}+ MgO ---> Fe_{2}O_{3}+ 3MgCl_{2}The Fe also gets balanced in this step.

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

2FeCl_{3}+ 3MgO ---> Fe_{2}O_{3}+ 3MgCl_{2}The other element (Mg or O, depending on which one you picked) also gets balanced in this step.

**Problem #2:** Li + H_{3}PO_{4} ---> H_{2} + Li_{3}PO_{4}

**Solution:**

1) Balance the Li:

3Li + H_{3}PO_{4}---> H_{2}+ Li_{3}PO_{4}

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 + 2H_{3}PO_{4}---> 3H_{2}+ Li_{3}PO_{4}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 + 2H_{3}PO_{4}---> 3H_{2}+ 2Li_{3}PO_{4}

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

6Li + 2H_{3}PO_{4}---> 3H_{2}+ 2Li_{3}PO_{4}

**Problem #3:** ZnS + O_{2} ---> ZnO + SO_{2}

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

ZnS + (3/2)O_{2}---> ZnO + SO_{2}

2) Multiply through to clear the fraction:

2ZnS + 3O_{2}---> 2ZnO + 2SO_{2}

**Problem #4:** FeS_{2} + Cl_{2} ---> FeCl_{3} + S_{2}Cl_{2}

**Solution:**

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:

FeS_{2}+ (5/2)Cl_{2}---> FeCl_{3}+ S_{2}Cl_{2}

Clear the fraction by multiplying through by 2:

2FeS_{2}+ 5Cl_{2}---> 2FeCl_{3}+ 2S_{2}Cl_{2}

**Problem #5:** Fe + HC_{2}H_{3}O_{2} ---> Fe(C_{2}H_{3}O_{2})_{3} + H_{2}

**Solution:**

1) Balance the acetate:

Fe + 3HC_{2}H_{3}O_{2}---> Fe(C_{2}H_{3}O_{2})_{3}+ H_{2}

2) Balance the hydrogen:

Fe + 3HC_{2}H_{3}O_{2}---> Fe(C_{2}H_{3}O_{2})_{3}+ (3/2)H_{2}

3) Clear the fraction:

2Fe + 6HC_{2}H_{3}O_{2}---> 2Fe(C_{2}H_{3}O_{2})_{3}+ 3H_{2}

**Problem #6:** H_{2}(g) + V_{2}O_{5}(s) ---> V_{2}O_{3}(s) + H_{2}O(ℓ)

**Solution:**

1) Balance the oxygen:

H_{2}(g) + V_{2}O_{5}(s) ---> V_{2}O_{3}(s) + 2H_{2}O(ℓ)

2) Balance the hydrogen:

2H_{2}(g) + V_{2}O_{5}(s) ---> V_{2}O_{3}(s) + 2H_{2}O(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) + MnO_{2}(s) ---> MnCl_{2}(aq) + Cl_{2}(g) + H_{2}O(ℓ)

**Solution:**

1) Balance the chlorine:

4HCl(aq) + MnO_{2}(s) ---> MnCl_{2}(aq) + Cl_{2}(g) + H_{2}O(ℓ)

2) Balance the hydrogen:

4HCl(aq) + MnO_{2}(s) ---> MnCl_{2}(aq) + Cl_{2}(g) + 2H_{2}O(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:** Fe_{2}O_{3}(s) + C(s) ---> Fe(s) + CO_{2}(g)

**Solution:**

1) Balance the iron:

Fe_{2}O_{3}(s) + C(s) ---> 2Fe(s) + CO_{2}(g)

2) Balance the oxygen:

Fe_{2}O_{3}(s) + C(s) ---> 2Fe(s) + (3/2)CO_{2}(g)

3) Balance the carbon:

Fe_{2}O_{3}(s) + (3/2)C(s) ---> 2Fe(s) + (3/2)CO_{2}(g)

Note the three-halves in front of the C and the CO_{2}. What's that you say? You can't have a three-halves of an atom? Ah, just you wait.

4) Multiply through by two for the final answer:

2Fe_{2}O_{3}(s) + 3C(s) ---> 4Fe(s) + 3CO_{2}(g)

Comment: one way to look at this is that using the three-halves 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.

**Problem #9:** C_{5}H_{11}NH_{2} + O_{2} ---> CO_{2} + H_{2}O + NO_{2}

**Solution:**

1) Balance the hydrogens first:

2C_{5}H_{11}NH_{2}+ O_{2}---> CO_{2}+ 13H_{2}O + NO_{2}

Notice that I used a 2 in front of C_{5}H_{11}NH_{2}. That's because I knew that there are 13 hydrogens in the C_{5}H_{11}NH_{2} 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:

2C_{5}H_{11}NH_{2}+ O_{2}---> 10CO_{2}+ 13H_{2}O + 2NO_{2}

3) Oxygen:

2C_{5}H_{11}NH_{2}+ (37/2)O_{2}---> 10CO_{2}+ 13H_{2}O + 2NO_{2}

4) Multiply through by 2 for:

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

**Problem #10:** CO_{2} + S_{8} ---> CS_{2} + SO_{2}

**Solution:**

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

CO_{2}+ (3/8)S_{8}---> CS_{2}+ SO_{2}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. Not too often does one see 3/8. Pretty tricky!

2) Multiply through by 8:

8CO_{2}+ 3S_{8}---> 8CS_{2}+ 8SO_{2}

**Bonus Problem:** P_{4} + O_{2} ---> P_{2}O_{3}

**Solution:**

1) Suppose you decide to balance the oxygen first:

P_{4}+ 3O_{2}---> 2P_{2}O_{3}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):

P_{4}+ O_{2}---> 2P_{2}O_{3}Then, the oxygen gets balanced:

P

_{4}+ 3O_{2}---> 2P_{2}O_{3}

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

(1/2)P_{4}+ O_{2}---> P_{2}O_{3}Let us balance the oxygen with a fraction as well:

(1/2)P

_{4}+ (3/2)O_{2}---> P_{2}O_{3}Finally, multiply through by two:

P

_{4}+ 3O_{2}---> 2P_{2}O_{3}

Balancing equations is fun!!

Fifteen examples | Problems 11-25 | Problems 26-45 | Problems 46-65 |

Six "balancing by groups" problems | Only the problems | Return to Equations Menu | |

Sixteen balance redox equations by sight |