Problems #46 - 65

Fifteen examples | Problems 1-10 | Problems 11-25 | Problems 26-45 |

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

Sixteen balance redox equations by sight |

**Problem #46:** NaBH_{4} + BF_{3} ---> NaBF_{4} + B_{2}H_{6}

**Solution:**

1) Balance the F:

NaBH_{4}+ 4BF_{3}---> 3NaBF_{4}+ B_{2}H_{6}

2) Balance the H:

3NaBH_{4}+ 4BF_{3}---> 3NaBF_{4}+ 2B_{2}H_{6}Notice that the Na is also balanced. Count up the B to make sure it is balanced also.

Why did I do what I did? Note that the F only occurs in one compound on the left and one compound on the right, so I ignored the boron and concentrated on the fluorine.

Then, I did the same thing for the H. It's in only one compound on the left as well as only one compound on the right. So, once again, I ignored everything else and balanced the H. Then, I checked the Na and the B and determined that the equation was now balanced.

**Problem #47:** C_{5}H_{7}O + O_{2} ---> CO_{2} + H_{2}O

**Solution:**

1) Balance carbon:

C_{5}H_{7}O + O_{2}---> 5CO_{2}+ H_{2}O

2) Balance hydrogen:

2C_{5}H_{7}O + O_{2}---> 10CO_{2}+ 7H_{2}ONote that I also took the carbon on the right to 10 as a consequence of putting the 2 in front of C

_{5}H_{7}O.

3) Balance oxygen:

2C_{5}H_{7}O +^{25}⁄_{2}O_{2}---> 10CO_{2}+ 7H_{2}O

4) Clear fraction:

4C_{5}H_{7}O + 25O_{2}---> 20CO_{2}+ 14H_{2}O

Balancing the hydrogens in step two was interesting due to the seven H present on the left-hand side. The problem is that the hydrogens only come in even numbers of the right-hand side, due to the H_{2}O. So, the solution is to make an even number of hydrogens on the left-hand side with the coefficient of two in front of the C_{5}H_{7}O.

**Problem #48:** NaOH + P_{4} + H_{2}O ---> NaH_{2}PO_{2} + PH_{3}

**Solution:**

1) Phosphorus:

NaOH + P_{4}+ H_{2}O ---> 3NaH_{2}PO_{2}+ PH_{3}

I picked using the P in the NaH_{2}PO_{2} rather than the PH_{3} because I knew I'd have to do both H and O. Putting a 3 in front of the NaH_{2}PO_{2} gives me more O to work with.

2) Sodium:

3NaOH + P_{4}+ H_{2}O ---> 3NaH_{2}PO_{2}+ PH_{3}

3) Balance the H and the O at the same time:

3NaOH + P_{4}+ 3H_{2}O ---> 3NaH_{2}PO_{2}+ PH_{3}

**Problem #49:** NBr_{3} + NaOH ----> N_{2} + NaBr + HOBr

**Solution:**

1) Nitrogen:

2NBr_{3}+ NaOH ----> N_{2}+ NaBr + HOBr

Some explanation: the next step is to balance the Br, however there are issues. Notice how the Na and the OH are both in one compound on the left, but are in TWO compounds on the right. The consequence of that is that the coefficient in front of the NaBr must be the same as the coefficient in front of the HOBr. At the same time, those two coefficients MUST provide six bromines.

2) That being said, here's the next step:

2NBr_{3}+ NaOH ----> N_{2}+ 3NaBr + 3HOBr

And that's the answer.

**Problem #50:** VO + Fe_{2}O_{3} ---> FeO + V_{2}O_{5}

**Solution:**

The oxygen issue is complex, it being in every compound in the problem. However, let us try to balance the V and the Fe and see what happens.

1) Balance the V:

2VO + Fe_{2}O_{3}---> FeO + V_{2}O_{5}

2) Balance the Fe:

2VO + Fe_{2}O_{3}---> 2FeO + V_{2}O_{5}

This does not balance the oxygens.

3) Let us not give up hope. Double the coefficients in front of the Fe, going from 1, 2 to 2, 4:

2VO + 2Fe_{2}O_{3}---> 4FeO + V_{2}O_{5}

This does not balance the oxygens. However, something has happened. The gap in the oxygens has closed. With Fe coefficients of 1, 2, there was a two oxygen gap. With Fe coefficients of 2, 4; there is a one oxygen gap. Let us continue.

4) Try a new set of coefficients (3, 6) in front of the Fe:

2VO + 3Fe_{2}O_{3}---> 6FeO + V_{2}O_{5}

It's balanced.

Note: I was once asked this question:

"How did you know what coefficients to add to the Fe to balance the O, and why did you focus on the Fe as opposed to the V?"

Look at just the Fe in step 2 and see that the coefficients are 1 and 2. Now, look at the Fe coefficients in step 3. All I did was try the next set of Fe coefficients (the 2 and the 4) that kept the Fe balanced. So, after noticing that the oxygen gap has closed by 1 (when I changed the Fe coefficients), I simply went to the next set of coefficients (the 3 and the 6) that kept the Fe balanced. And the oxygen gap closed by one more, balancing the oxygens.

I selected the Fe to do this to rather than the V because closing the oxygen gap does not happen with the V. In step 2, if I change the V rather than the Fe, I get this:

4VO + Fe_{2}O_{3}---> 2FeO + 2V_{2}O_{5}

and the oxygen difference increases from 2 (in step 2) to 5 (in the equation just above. If I try V, I make my problem worse, so I try the Fe and eventually have success there.

**Problem #51:** Fe_{2}O_{3}(s) + CO(g) ---> Fe(s) + CO_{2}(g)

**Solution:**

1) Balance the Fe:

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

To balance the oxygen, we follow the same path as balancing the oxygens in the problem just above. Notice that the carbon is balanced. What we will do is try new coefficients that keep the carbon balanced.

2) Trial #1:

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

This does not balance the oxygens, but something good does happen. When the carbon coefficients were 1 & 1, there was a two oxygen gap. With carbon coefficients of 2 & 2, the gap is now only one oxygen.

3) Trial #2:

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

It's balanced.

**Problem #52:** S_{2}H_{5} + O_{2} ---> SO_{2} + H_{2}O

**Solution:**

1) Balance the H:

2S_{2}H_{5}+ O_{2}---> SO_{2}+ 5H_{2}ONotice how I did it with two coefficients at the same time. That is because 10 is the least common multiple between 2 and 5.

2) Balance the S:

2S_{2}H_{5}+ O_{2}---> 4SO_{2}+ 5H_{2}O

3) Balance the O:

2S_{2}H_{5}+^{13}⁄_{2}O_{2}---> 4SO_{2}+ 5H_{2}O

4) Clear the fraction:

4S_{2}H_{5}+ 13O_{2}---> 8SO_{2}+ 10H_{2}O

**Problem #53:** KOH + F_{2} ---> KF + F_{2}O + H_{2}O

**Solution:**

1) Balance the H:

2KOH + F_{2}---> KF + F_{2}O + H_{2}OThis also balances the oxygen.

2) Balance the K:

2KOH + F_{2}---> 2KF + F_{2}O + H_{2}O

3) Balance the F:

2KOH + 2F_{2}---> 2KF + F_{2}O + H_{2}O

**Problem #54:** Al + KOH + H_{2}SO_{4} + H_{2}O ----> KAl(SO_{4})_{2} **·** 12H_{2}O + H_{2}

**Solution #1 - keep all water (solution below removes the water, then balances):**

1) Balance the water of hydration:

Al + KOH + H_{2}SO_{4}+ 12H_{2}O ----> KAl(SO_{4})_{2}·12H_{2}O + H_{2}

2) Balance the sulfates:

Al + KOH + 2H_{2}SO_{4}+ 12H_{2}O ----> KAl(SO_{4})_{2}·12H_{2}O + H_{2}

3) Notice that I reduce the water on the left by one:

Al + KOH + 2H_{2}SO_{4}+ 11H_{2}O ----> KAl(SO_{4})_{2}·12H_{2}O + H_{2}

I'm doing this because I still need to take care of the hydroxide and the way to do that is by adding a hydrogen to it and make water. This water will replace the one water that I eliminated from the left-hand side.

4) Balance the hydrogens:

Al + KOH + 2H_{2}SO_{4}+ 11H_{2}O ----> KAl(SO_{4})_{2}·12H_{2}O + (3/2)H_{2}

There were 4 hydrogens in the sulfuric acid that had to be balanced. One of the H reacts with the hydroxide to replace the twelfth water I eliminated on the left-hand side above. The 3/2 on the right balances the other three that are left over.

5) Multiply through by two to clear the fractional coefficient:

2Al + 2KOH + 4H_{2}SO_{4}+ 22H_{2}O ----> 2KAl(SO_{4})_{2}·12H_{2}O + 3H_{2}

**Solution #2 - remove all water, add it back in at the end:**

1) Eliminate water:

Al + KOH + H_{2}SO_{4}----> KAl(SO_{4})_{2}+ H_{2}

2) Balance sulfate:

Al + KOH + 2H_{2}SO_{4}----> KAl(SO_{4})_{2}+ H_{2}

3) React one hydrogen with the one hydroxide to make water:

Al + KOH + 2H_{2}SO_{4}----> KAl(SO_{4})_{2}+ H_{2}+ H_{2}O

Notice I added water in on the right-hand side and that it was on the left-hand side in the original formulation of the problem. Be patient!

4) Balance the three remaining hydrogens:

Al + KOH + 2H_{2}SO_{4}----> KAl(SO_{4})_{2}+ (3/2)H_{2}+ H_{2}O

5) Clear the fraction:

2Al + 2KOH + 4H_{2}SO_{4}----> 2KAl(SO_{4})_{2}+ 3H_{2}+ 2H_{2}O

6) Add back in the water of hydration:

2Al + 2KOH + 4H_{2}SO_{4}+ 22H_{2}O ----> 2KAl(SO_{4})_{2}·12H_{2}O + 3H_{2}

Notice that the two waters on the right-hand side are incorporated into the water of hydration and that an additional 22 waters are added in order to reach the 12 waters required for each (of the two) KAl(SO_{4})_{2}.

**Problem #55:** KAl(SO_{4})_{2} **·** 12H_{2}O + BaCl_{2} ---> KCl + AlCl_{3} + BaSO_{4} + H_{2}O

**Solution:**

Comment: remove the water of hydration before balancing. This reaction cannot happen unless the reactants are in solution, so the water of hydration won't appear in the reaction because it's mixed in with the solvent.

1) Remove the waters of hydration:

KAl(SO_{4})_{2}+ BaCl_{2}---> KCl + AlCl_{3}+ BaSO_{4}

2) Balance:

3) Add back in the waters of hydration, if so desired (or required to by your teacher):sulfate:

KAl(SO_{4})_{2}+ BaCl_{2}---> KCl + AlCl_{3}+ 2BaSO_{4}barium:

KAl(SO_{4})_{2}+ 2BaCl_{2}---> KCl + AlCl_{3}+ 2BaSO_{4}and it's balanced.

KAl(SO_{4})_{2}·12H_{2}O + 2BaCl_{2}---> KCl + AlCl_{3}+ 2BaSO_{4}+ 12H_{2}O

**Problem #56:** C_{10}H_{22} ---> C_{4}H_{10} + C_{2}H_{4}

**Solution:**

1) via trial and error:

Try a 2 in front of C_{4}H_{10}to balance the C. What effect on the H? Answer: H does not balance.Try a 3 in front of the C

_{2}H_{4}to balance the C. What effect on the H? Answer: H does balance.

2) via simultaneous equations:

C_{10}H_{22}---> xC_{4}H_{10}+ yC_{2}H_{4}10 = 4x + 2y

22 = 10x + 4y5 = 2x + y

11 = 5x + 2yy = 5 - 2x

11 = 5x + 2(5 - 2x)

11 = x + 10

x = 1

22 = (10) (1) + 4y

y = 3

**Problem #57:** Here is an already balanced equation:

NH_{2}CH_{2}COOH(s) +^{9}⁄_{4}O_{2}(g) ---> 2CO_{2}(g) +^{5}⁄_{2}H_{2}O(ℓ) +^{1}⁄_{2}N_{2}(g)

How was it arrived at?

**Solution:**

Let's balance the equation step by step, aiming to reproduce the balanced equation above:

1) Balance the N

NH_{2}CH_{2}COOH(s) + O_{2}(g) ---> CO_{2}(g) + H_{2}O(ℓ) +^{1}⁄_{2}N_{2}(g)

2) Balance the C

NH_{2}CH_{2}COOH(s) + O_{2}(g) ---> 2CO_{2}(g) + H_{2}O(ℓ) +^{1}⁄_{2}N_{2}(g)

3) Balance the H:

NH_{2}CH_{2}COOH(s) + O_{2}(g) ---> 2CO_{2}(g) +^{5}⁄_{2}H_{2}O(ℓ) +^{1}⁄_{2}N_{2}(g)

4) The above is the interesting step since we now have ^{5}⁄_{2}O resulting from the ^{5}⁄_{2}H_{2}O. That means the total oxygen on the right is now 4 from the CO_{2} and ^{5}⁄_{2} from the H_{2}O. Let's do this:

^{8}⁄_{2}O +^{5}⁄_{2}O =^{13}⁄_{2}O

5) Take away the two O in the NH_{2}CH_{2}COOH and you're left with ^{9}⁄_{2}O, so we need ^{9}⁄_{4} in front of the O_{2}:

NH_{2}CH_{2}COOH(s) +^{9}⁄_{4}O_{2}(g) ---> 2CO_{2}(g) +^{5}⁄_{2}H_{2}O(ℓ) +^{1}⁄_{2}N_{2}(g)You use

^{9}⁄_{4}because^{9}⁄_{4}times 2 equals^{9}⁄_{2}, the number of oxygens we needed.The final step would be to clear all fractions by multiplying through by 4.

The question above was answered sometime in 2011 on Yahoo Answers.. After giving the solution I reproduced above (with a minor amount of commentary I added), the answerer goes on to discuss an alternate way to balance the equation. It's a good suggestion, one where you anticipate the presence of ^{1}⁄_{2}N_{2} and start with a 2 rather than a 1 in front of the NH_{2}CH_{2}COOH

**Problem #58:** C_{3}H_{5}O_{9}N_{3} ---> CO_{2} + N_{2} + O_{2} + H_{2}O

**Solution:**

Before putting down any coefficients, examine the hydrogen. You will need a ^{5}⁄_{2} in front of the H_{2}O to balance the hydrogens on the left. However, this gives you ^{5}⁄_{2} oxygen atoms. While you can balance it with ^{5}⁄_{2} oxygen atoms, let's not do that.

1) That means this for a first step:

2C_{3}H_{5}O_{9}N_{3}---> CO_{2}+ N_{2}+ O_{2}+ H_{2}O

2) Balance everything but the oxygen:

2C_{3}H_{5}O_{9}N_{3}---> 6CO_{2}+ 3N_{2}+ O_{2}+ 5H_{2}O

3) We need 18 oxygens on the right-hand side:

2C_{3}H_{5}O_{9}N_{3}---> 6CO_{2}+ 3N_{2}+^{1}⁄_{2}O_{2}+ 5H_{2}O

4) Clear the fraction:

4C_{3}H_{5}O_{9}N_{3}---> 12CO_{2}+ 6N_{2}+ O_{2}+ 10H_{2}O

5) Let us insist on using ^{5}⁄_{2} in front of the H_{2}O. Here is everything balanced but the oxygen:

C_{3}H_{5}O_{9}N_{3}---> 3CO_{2}+^{3}⁄_{2}N_{2}+ O_{2}+^{5}⁄_{2}H_{2}O

6) Balance the O:

C_{3}H_{5}O_{9}N_{3}---> 3CO_{2}+^{3}⁄_{2}N_{2}+^{1}⁄_{4}O_{2}+^{5}⁄_{2}H_{2}OI used

^{1}⁄_{4}in front of the O_{2}because I only needed^{1}⁄_{2}more oxygen to balance the 9 oxygens that are on the left. On the right-hand side there were 6 oxygens from the CO_{2}and^{5}⁄_{2}oxygens from the H_{2}O. That made 8^{1}⁄_{2}oxygens, so I only needed^{1}⁄_{2}more oxygen.

7) Clear the fractions by multiplying through by 4:

4C_{3}H_{5}O_{9}N_{3}---> 12CO_{2}+ 6N_{2}+ O_{2}+ 10H_{2}O

**Problem #59:** C_{20}H_{45} + O_{2} ---> CO_{2} + H_{2}O

**Solution:**

1) Balance C:

C_{20}H_{45}+ O_{2}---> 20CO_{2}+ H_{2}O

2) Balance H:

oops, let's not do that, so do this:2C

_{20}H_{45}+ O_{2}---> 40CO_{2}+ H_{2}O <--- I doubled the amount of C and left everything else unbalancedComment: I choose not to do it because I would have to do this next:

^{45}⁄_{2}H_{2}Othere's nothing the matter with the

^{45}⁄_{2}O that results, but I'm choosing to avoid it simply because the equation is a bit easier to balance when you anticipate the^{45}⁄_{2}O and avoid it.

3) Now, balance H:

2C_{20}H_{45}+ O_{2}---> 40CO_{2}+ 45H_{2}O

4) Balance O:

2C_{20}H_{45}+^{125}⁄_{2}O_{2}---> 40CO_{2}+ 45H_{2}O

5) Multiply through by 2:

4C_{20}H_{45}+ 125O_{2}---> 80CO_{2}+ 90H_{2}O

6) Suppose you absolutely insisted on writing ^{45}⁄_{2}H_{2}O. You could still balance the equation.

**Solution #1:**

C_{20}H_{45}+ O_{2}---> 20CO_{2}+^{45}⁄_{2}H_{2}Oon the right, we have this:

^{45}⁄_{2}+ 40 oxygen atoms.that adds to

^{45}⁄_{2}+^{80}⁄_{2}=^{125}⁄_{2}O atoms.that means we need a coefficient of

^{125}⁄_{4}in front of the O_{2}.Why? We need

^{125}⁄_{2}O atoms and, on the left-hand side, they only come two at a time. So, the coefficient in front of the O_{2}needs to be half of the^{125}⁄_{2}.^{125}⁄_{2}divided by 2 equals^{125}⁄_{4}.C

_{20}H_{45}+^{125}⁄_{4}O_{2}---> 20CO_{2}+^{45}⁄_{2}H_{2}OMultiply through by 4 to get the final answer.

**Solution #2:**

Start here:

C_{20}H_{45}+ O_{2}---> 20CO_{2}+^{45}⁄_{2}H_{2}O

and multiply through by 2:

2C_{20}H_{45}+ 2O_{2}---> 40CO_{2}+ 45H_{2}O

Balance the oxygens:

2C_{20}H_{45}+^{125}⁄_{2}O_{2}---> 40CO_{2}+ 45H_{2}O

Multiply through by 2 to clear the fraction and you're done. Note that we did two "multiply by 2" as opposed to a single "multiply by 4" up above.

**Problem #60:** C_{3}H_{5} + O_{2} ---> CO_{2} + H_{2}O

**Solution #1:**

1) Balance C:

C_{3}H_{5}+ O_{2}---> 3CO_{2}+ H_{2}O

2) If I use ^{5}⁄_{2} to balance the H, like this:

C_{3}H_{5}+ O_{2}---> 3CO_{2}+^{5}⁄_{2}H_{2}OThat means I have

^{5}⁄_{2}O. To not have that, I do the following.

3) Go back and double step 1:

2C_{3}H_{5}+ O_{2}---> 6CO_{2}+ H_{2}O

4) Now, re-do step 2 and balance the hydrogen:

2C_{3}H_{5}+ O_{2}---> 6CO_{2}+ 5H_{2}O

5) Balance the O:

2C_{3}H_{5}+^{17}⁄_{2}O_{2}---> 6CO_{2}+ 5H_{2}O

6) Multiply through by 2:

4C_{3}H_{5}+ 17O_{2}---> 12CO_{2}+ 10H_{2}O

**Solution #2:**

1) Balance C:

C_{3}H_{5}+ O_{2}---> 3CO_{2}+ H_{2}O

2) Use ^{5}⁄_{2} to balance the H:

C_{3}H_{5}+ O_{2}---> 3CO_{2}+^{5}⁄_{2}H_{2}O

3) Double the coefficients from step 2:

2C_{3}H_{5}+ 2O_{2}---> 6CO_{2}+ 5H_{2}O

4) Balance the oxygens:

2C_{3}H_{5}+^{17}⁄_{2}O_{2}---> 6CO_{2}+ 5H_{2}OTo do the above, I simply added up the O on the right and get 17. I then replaced the 2 in front of the O

_{2}with^{17}⁄_{2}.

5) Multiply by 2:

4C_{3}H_{5}+ 17O_{2}---> 12CO_{2}+ 10H_{2}O

Comment: given ^{5}⁄_{2}H_{2}O, you could also balance the oxygen with ^{17}⁄_{4}O_{2}. You then multiply through by a factor of 4 to get the final answer.

**Problem #61:** C_{5}H_{11}SO_{2}N + O_{2} ---> CO_{2} + H_{2}O + SO_{2} + NO_{2}

**Solution:**

1) If we proceed with the normal balancing pattern, we would have 5CO_{2} and then we would have ^{11}⁄_{2}H_{2}O, the last of which gives up eleven hydrogens, which is what we want. We would also have ^{11}⁄_{2} oxygens. This isn't as big a problem as it seems, but let's avoid it this way:

2C_{5}H_{11}SO_{2}N + O_{2}---> CO_{2}+ H_{2}O + SO_{2}+ NO_{2}

2) Balance the carbon:

2C_{5}H_{11}SO_{2}N + O_{2}---> 10CO_{2}+ H_{2}O + SO_{2}+ NO_{2}

3) Balance the hydrogen:

2C_{5}H_{11}SO_{2}N + O_{2}---> 10CO_{2}+ 11H_{2}O + SO_{2}+ NO_{2}

4) Balance the sulfur and the nitrogen:

2C_{5}H_{11}SO_{2}N + O_{2}---> 10CO_{2}+ 11H_{2}O + 2SO_{2}+ 2NO_{2}

5) Balance the oxygen:

There are 39 O on the right and there are already 4 O present on the left. So:2C

_{5}H_{11}SO_{2}N +^{35}⁄_{2}O_{2}---> 10CO_{2}+ 11H_{2}O + 2SO_{2}+ 2NO_{2}

6) Multiply through by 2 to clear the fraction:

4C_{5}H_{11}SO_{2}N + 35O_{2}---> 20CO_{2}+ 22H_{2}O + 4SO_{2}+ 4NO_{2}

7) Suppose you insisted on using ^{11}⁄_{2} in front of the H_{2}O. We could still balance it:

C_{5}H_{11}SO_{2}N + O_{2}---> 5CO_{2}+^{11}⁄_{2}H_{2}O + SO_{2}+ NO_{2}Count the oxygens on the right and get a total of

^{39}⁄_{2}. There are 2 O already present on the left side, so we need^{35}⁄_{2}more oxygen. Put a^{35}⁄_{4}in front of the O_{2}because^{35}⁄_{4}times 2 equals^{35}⁄_{2}.C

_{5}H_{11}SO_{2}N +^{35}⁄_{4}O_{2}---> 5CO_{2}+^{11}⁄_{2}H_{2}O + SO_{2}+ NO_{2}Multiply through by 4 to get the final answer.

**Problem #62:** Ca(OH)_{2} + C_{4}H_{6}O_{5} ---> H_{2}O + CaC_{4}H_{4}O_{5}

**Solution:**

Ignore the Ca and the C_{4}__O_{5}.

Instead, look only at this:

(OH)_{2}+ H_{6}---> H_{2}O + H_{4}

And balance:

(OH)_{2}+ H_{6}---> 2H_{2}O + H_{4}

And the answer is:

Ca(OH)_{2}+ C_{4}H_{6}O_{5}---> 2H_{2}O + CaC_{4}H_{4}O_{5}

We ignored the portion of the equation that was already balanced.

**Problem #63:** FeTiO_{3} + Cl_{2} + C ---> TiCl_{4} + FeCl_{3} + CO

**Solution:**

1) Balance Cl:

FeTiO_{3}+^{7}⁄_{2}Cl_{2}+ C ---> TiCl_{4}+ FeCl_{3}+ CO

2) Balance O:

FeTiO_{3}+^{7}⁄_{2}Cl_{2}+ C ---> TiCl_{4}+ FeCl_{3}+ 3CO

3) Balance C:

FeTiO_{3}+^{7}⁄_{2}Cl_{2}+ 3C ---> TiCl_{4}+ FeCl_{3}+ 3CO

4) Multiply through by 2:

2FeTiO_{3}+ 7Cl_{2}+ 6C ---> 2TiCl_{4}+ 2FeCl_{3}+ 6CO

Note: You could have started the balancing with the O, then gone to the C and follow that with the Cl. You would have wound up in the same place as above.

**Problem #64:** Ca_{3}P_{2}(s) + H_{2}O(ℓ) ---> Ca(OH)_{2}(aq) + PH_{3}(g)

**Solution:**

1) Balance the Ca:

Ca_{3}P_{2}(s) + H_{2}O(ℓ) ---> 3Ca(OH)_{2}(aq) + PH_{3}(g)

2) Balance the P:

Ca_{3}P_{2}(s) + H_{2}O(ℓ) ---> 3Ca(OH)_{2}(aq) + 2PH_{3}(g)

3) Balance the OH and the H at the same time:

Ca_{3}P_{2}(s) + 6H_{2}O(ℓ) ---> 3Ca(OH)_{2}(aq) + 2PH_{3}(g)

Note that there are 6 OH and 6 H on the right-hand side, which is why I used a 6 in front of the H_{2}O on the left-hand side.

**Problem #65:** TiCl_{4}(ℓ) + H_{2}O(ℓ) ---> TiO_{2}(s) + HCl(aq)

1) Look at the Ti:

TiCl_{4}(ℓ) + H_{2}O(ℓ) ---> TiO_{2}(s) + HCl(aq)One on each side, it's balanced.

2) Look at the Cl. 4 on the left, so we need 4 on the right:

TiCl_{4}(ℓ) + H_{2}O(ℓ) ---> TiO_{2}(s) + 4HCl(aq)

3) Look at the H. 4 on the right, because of the 4HCl, and 2 on the left.

TiCl_{4}(ℓ) + 2H_{2}O(ℓ) ---> TiO_{2}(s) + 4HCl(aq)The 2 in front of the H

_{2}O balances the H and it also balances the O.

Fifteen examples | Problems 1-10 | Problems 11-25 | Problems 26-45 |

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

Sixteen balance redox equations by sight |