### Balancing Chemical EquationsProblems #46 - 65

 Twenty examples Probs 1-10 Probs 11-25 Probs 26-45 "Balancing by groups" problems Only the problems Return to Equations Menu Balance redox equations by sight

Problem #46: NaBH4 + BF3 ---> NaBF4 + B2H6

Solution:

1) Balance the F:

NaBH4 + 4BF3 ---> 3NaBF4 + B2H6

2) Balance the H:

3NaBH4 + 4BF3 ---> 3NaBF4 + 2B2H6

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: C5H7O + O2 ---> CO2 + H2O

Solution:

1) Balance carbon:

C5H7O + O2 ---> 5CO2 + H2O

2) Balance hydrogen:

2C5H7O + O2 ---> 10CO2 + 7H2O

Note that I also took the carbon on the right to 10 as a consequence of putting the 2 in front of C5H7O.

3) Balance oxygen:

2C5H7O + 252O2 ---> 10CO2 + 7H2O

4) Clear fraction:

4C5H7O + 25O2 ---> 20CO2 + 14H2O

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 H2O. 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 C5H7O.

Problem #48: NaOH + P4 + H2O ---> NaH2PO2 + PH3

Solution:

1) Phosphorus:

NaOH + P4 + H2O ---> 3NaH2PO2 + PH3

I picked using the P in the NaH2PO2 rather than the PH3 because I knew I'd have to do both H and O. Putting a 3 in front of the NaH2PO2 gives me more O to work with.

2) Sodium:

3NaOH + P4 + H2O ---> 3NaH2PO2 + PH3

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

3NaOH + P4 + 3H2O ---> 3NaH2PO2 + PH3

Problem #49: NBr3 + NaOH ----> N2 + NaBr + HOBr

Solution:

1) Nitrogen:

2NBr3 + NaOH ----> N2 + 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:

2NBr3 + NaOH ----> N2 + 3NaBr + 3HOBr

Problem #50: VO + Fe2O3 ---> FeO + V2O5

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 + Fe2O3 ---> FeO + V2O5

2) Balance the Fe:

2VO + Fe2O3 ---> 2FeO + V2O5

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 + 2Fe2O3 ---> 4FeO + V2O5

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 + 3Fe2O3 ---> 6FeO + V2O5

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 + Fe2O3 ---> 2FeO + 2V2O5

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: Fe2O3(s) + CO(g) ---> Fe(s) + CO2(g)

Solution:

1) Balance the Fe:

Fe2O3(s) + CO(g) ---> 2Fe(s) + CO2(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:

Fe2O3(s) + 2CO(g) ---> 2Fe(s) + 2CO2(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:

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

It's balanced.

Problem #52: S2H5 + O2 ---> SO2 + H2O

Solution:

1) Balance the H:

2S2H5 + O2 ---> SO2 + 5H2O

Notice 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:

2S2H5 + O2 ---> 4SO2 + 5H2O

3) Balance the O:

2S2H5 + 132O2 ---> 4SO2 + 5H2O

4) Clear the fraction:

4S2H5 + 13O2 ---> 8SO2 + 10H2O

Problem #53: KOH + F2 ---> KF + F2O + H2O

Solution:

1) Balance the H:

2KOH + F2 ---> KF + F2O + H2O

This also balances the oxygen.

2) Balance the K:

2KOH + F2 ---> 2KF + F2O + H2O

3) Balance the F:

2KOH + 2F2 ---> 2KF + F2O + H2O

Problem #54: Al + KOH + H2SO4 + H2O ----> KAl(SO4)2 · 12H2O + H2

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

1) Balance the water of hydration:

Al + KOH + H2SO4 + 12H2O ----> KAl(SO4)2 · 12H2O + H2

2) Balance the sulfates:

Al + KOH + 2H2SO4 + 12H2O ----> KAl(SO4)2 · 12H2O + H2

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

Al + KOH + 2H2SO4 + 11H2O ----> KAl(SO4)2 · 12H2O + H2

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 + 2H2SO4 + 11H2O ----> KAl(SO4)2 · 12H2O + (3/2)H2

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 + 4H2SO4 + 22H2O ----> 2KAl(SO4)2 · 12H2O + 3H2

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

1) Eliminate water:

Al + KOH + H2SO4 ----> KAl(SO4)2 + H2

2) Balance sulfate:

Al + KOH + 2H2SO4 ----> KAl(SO4)2 + H2

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

Al + KOH + 2H2SO4 ----> KAl(SO4)2 + H2 + H2O

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 + 2H2SO4 ----> KAl(SO4)2 + (3/2)H2 + H2O

5) Clear the fraction:

2Al + 2KOH + 4H2SO4 ----> 2KAl(SO4)2 + 3H2 + 2H2O

6) Add back in the water of hydration:

2Al + 2KOH + 4H2SO4 + 22H2O ----> 2KAl(SO4)2 · 12H2O + 3H2

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(SO4)2.

Problem #55: KAl(SO4)2 · 12H2O + BaCl2 ---> KCl + AlCl3 + BaSO4 + H2O

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(SO4)2 + BaCl2 ---> KCl + AlCl3 + BaSO4

2) Balance:

sulfate:

KAl(SO4)2 + BaCl2 ---> KCl + AlCl3 + 2BaSO4

barium:

KAl(SO4)2 + 2BaCl2 ---> KCl + AlCl3 + 2BaSO4

and it's balanced.

3) Add back in the waters of hydration, if so desired (or required to by your teacher):
KAl(SO4)2 · 12H2O + 2BaCl2 ---> KCl + AlCl3 + 2BaSO4 + 12H2O

Problem #56: C10H22 ---> C4H10 + C2H4

Solution:

1) Via trial and error:

Try a 2 in front of C4H10 to balance the C. What effect on the H? Answer: H does not balance.

Try a 3 in front of the C2H4 to balance the C. What effect on the H? Answer: H does balance.

2) via simultaneous equations:

C10H22 ---> xC4H10 + yC2H4

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

5 = 2x + y
11 = 5x + 2y

y = 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:

NH2CH2COOH(s) + 94O2(g) ---> 2CO2(g) + 52H2O(ℓ) + 12N2(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

NH2CH2COOH(s) + O2(g) ---> CO2(g) + H2O(ℓ) + 12N2(g)

2) Balance the C

NH2CH2COOH(s) + O2(g) ---> 2CO2(g) + H2O(ℓ) + 12N2(g)

3) Balance the H:

NH2CH2COOH(s) + O2(g) ---> 2CO2(g) + 52H2O(ℓ) + 12N2(g)

4) The above is the interesting step since we now have 52O resulting from the 52H2O. That means the total oxygen on the right is now 4 from the CO2 and 52 from the H2O. Let's do this:

82O + 52O = 132O

5) Take away the two O in the NH2CH2COOH and you're left with 92O, so we need 94 in front of the O2:

NH2CH2COOH(s) + 94O2(g) ---> 2CO2(g) + 52H2O(ℓ) + 12N2(g)

You use 94 because 94 times 2 equals 92, 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 12N2 and start with a 2 rather than a 1 in front of the NH2CH2COOH

Problem #58: C3H5O9N3 ---> CO2 + N2 + O2 + H2O

Solution:

Before putting down any coefficients, examine the hydrogen. You will need a 52 in front of the H2O to balance the hydrogens on the left. However, this gives you 52 oxygen atoms. While you can balance it with 52 oxygen atoms, let's not do that.

1) That means this for a first step:

2C3H5O9N3 ---> CO2 + N2 + O2 + H2O

2) Balance everything but the oxygen:

2C3H5O9N3 ---> 6CO2 + 3N2 + O2 + 5H2O

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

2C3H5O9N3 ---> 6CO2 + 3N2 + 12O2 + 5H2O

4) Clear the fraction:

4C3H5O9N3 ---> 12CO2 + 6N2 + O2 + 10H2O

5) Let us insist on using 52 in front of the H2O. Here is everything balanced but the oxygen:

C3H5O9N3 ---> 3CO2 + 32N2 + O2 + 52H2O

6) Balance the O:

C3H5O9N3 ---> 3CO2 + 32N2 + 14O2 + 52H2O

I used 14 in front of the O2 because I only needed 12 more oxygen to balance the 9 oxygens that are on the left. On the right-hand side there were 6 oxygens from the CO2 and 52 oxygens from the H2O. That made 812 oxygens, so I only needed 12 more oxygen.

7) Clear the fractions by multiplying through by 4:

4C3H5O9N3 ---> 12CO2 + 6N2 + O2 + 10H2O

Problem #59: C20H45 + O2 ---> CO2 + H2O

Solution:

1) Balance C:

C20H45 + O2 ---> 20CO2 + H2O

2) Balance H:

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

2C20H45 + O2 ---> 40CO2 + H2O <--- I doubled the amount of C and left everything else unbalanced

Comment: I choose not to do it because I would have to do this next:

452H2O

there's nothing the matter with the 452O that results, but I'm choosing to avoid it simply because the equation is a bit easier to balance when you anticipate the 452O and avoid it.

3) Now, balance H:

2C20H45 + O2 ---> 40CO2 + 45H2O

4) Balance O:

2C20H45 + 1252O2 ---> 40CO2 + 45H2O

5) Multiply through by 2:

4C20H45 + 125O2 ---> 80CO2 + 90H2O

6) Suppose you absolutely insisted on writing 452H2O. You could still balance the equation.

Solution #1:

C20H45 + O2 ---> 20CO2 + 452H2O

on the right, we have this: 452 + 40 oxygen atoms.

that adds to 452 + 802 = 1252O atoms.

that means we need a coefficient of 1254 in front of the O2.

Why? We need 1252O atoms and, on the left-hand side, they only come two at a time. So, the coefficient in front of the O2 needs to be half of the 1252. 1252 divided by 2 equals 1254.

C20H45 + 1254O2 ---> 20CO2 + 452H2O

Multiply through by 4 to get the final answer.

Solution #2:

Start here:

C20H45 + O2 ---> 20CO2 + 452H2O

and multiply through by 2:

2C20H45 + 2O2 ---> 40CO2 + 45H2O

Balance the oxygens:

2C20H45 + 1252O2 ---> 40CO2 + 45H2O

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: C3H5 + O2 ---> CO2 + H2O

Solution #1:

1) Balance C:

C3H5 + O2 ---> 3CO2 + H2O

2) If I use 52 to balance the H, like this:

C3H5 + O2 ---> 3CO2 + 52H2O

That means I have 52O. To not have that, I do the following.

3) Go back and double step 1:

2C3H5 + O2 ---> 6CO2 + H2O

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

2C3H5 + O2 ---> 6CO2 + 5H2O

5) Balance the O:

2C3H5 + 172O2 ---> 6CO2 + 5H2O

6) Multiply through by 2:

4C3H5 + 17O2 ---> 12CO2 + 10H2O

Solution #2:

1) Balance C:

C3H5 + O2 ---> 3CO2 + H2O

2) Use 52 to balance the H:

C3H5 + O2 ---> 3CO2 + 52H2O

3) Double the coefficients from step 2:

2C3H5 + 2O2 ---> 6CO2 + 5H2O

4) Balance the oxygens:

2C3H5 + 172O2 ---> 6CO2 + 5H2O

To do the above, I simply added up the O on the right and get 17. I then replaced the 2 in front of the O2 with 172.

5) Multiply by 2:

4C3H5 + 17O2 ---> 12CO2 + 10H2O

Comment: given 52H2O, you could also balance the oxygen with 174O2. You then multiply through by a factor of 4 to get the final answer.

Problem #61: C5H11SO2N + O2 ---> CO2 + H2O + SO2 + NO2

Solution:

1) If we proceed with the normal balancing pattern, we would have 5CO2 and then we would have 112H2O, the last of which gives up eleven hydrogens, which is what we want. We would also have 112 oxygens. This isn't as big a problem as it seems, but let's avoid it this way:

2C5H11SO2N + O2 ---> CO2 + H2O + SO2 + NO2

2) Balance the carbon:

2C5H11SO2N + O2 ---> 10CO2 + H2O + SO2 + NO2

3) Balance the hydrogen:

2C5H11SO2N + O2 ---> 10CO2 + 11H2O + SO2 + NO2

4) Balance the sulfur and the nitrogen:

2C5H11SO2N + O2 ---> 10CO2 + 11H2O + 2SO2 + 2NO2

5) Balance the oxygen:

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

2C5H11SO2N + 352O2 ---> 10CO2 + 11H2O + 2SO2 + 2NO2

6) Multiply through by 2 to clear the fraction:

4C5H11SO2N + 35O2 ---> 20CO2 + 22H2O + 4SO2 + 4NO2

7) Suppose you insisted on using 112 in front of the H2O. We could still balance it:

C5H11SO2N + O2 ---> 5CO2 + 112H2O + SO2 + NO2

Count the oxygens on the right and get a total of 392. There are 2 O already present on the left side, so we need 352 more oxygen. Put a 354 in front of the O2 because 354 times 2 equals 352.

C5H11SO2N + 354O2 ---> 5CO2 + 112H2O + SO2 + NO2

Multiply through by 4 to get the final answer.

Problem #62: Ca(OH)2 + C4H6O5 ---> H2O + CaC4H4O5

Solution:

Ignore the Ca and the C4__O5.

(OH)2 + H6 ---> H2O + H4

And balance:

(OH)2 + H6 ---> 2H2O + H4

Ca(OH)2 + C4H6O5 ---> 2H2O + CaC4H4O5

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

Problem #63: FeTiO3 + Cl2 + C ---> TiCl4 + FeCl3 + CO

Solution:

1) Balance Cl:

FeTiO3 + 72Cl2 + C ---> TiCl4 + FeCl3 + CO

2) Balance O:

FeTiO3 + 72Cl2 + C ---> TiCl4 + FeCl3 + 3CO

3) Balance C:

FeTiO3 + 72Cl2 + 3C ---> TiCl4 + FeCl3 + 3CO

4) Multiply through by 2:

2FeTiO3 + 7Cl2 + 6C ---> 2TiCl4 + 2FeCl3 + 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: Ca3P2(s) + H2O(ℓ) ---> Ca(OH)2(aq) + PH3(g)

Solution:

1) Balance the Ca:

Ca3P2(s) + H2O(ℓ) ---> 3Ca(OH)2(aq) + PH3(g)

2) Balance the P:

Ca3P2(s) + H2O(ℓ) ---> 3Ca(OH)2(aq) + 2PH3(g)

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

Ca3P2(s) + 6H2O(ℓ) ---> 3Ca(OH)2(aq) + 2PH3(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 H2O on the left-hand side.

Problem #65: TiCl4(ℓ) + H2O(ℓ) ---> TiO2(s) + HCl(aq)

1) Look at the Ti:

TiCl4(ℓ) + H2O(ℓ) ---> TiO2(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:

TiCl4(ℓ) + H2O(ℓ) ---> TiO2(s) + 4HCl(aq)

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

TiCl4(ℓ) + 2H2O(ℓ) ---> TiO2(s) + 4HCl(aq)

The 2 in front of the H2O balances the H and it also balances the O.

Problem #66: Cu2O + O2 ---> CuO

1) Balance Cu:

Cu2O + O2 ---> 2CuO

2) Balance the O:

Cu2O + 12O2 ---> 2CuO

Notice how I used the 12 to take away an oxygen atom. Sometimes, you balance by taking away.

3) Multiply through by 2:

2Cu2O + O2 ---> 4CuO

Problem #67: FeSO4(NH4)2SO4 6H2O + H2C2O4 2H2O ---> FeC2O4 2H2O + NH4HSO4 + H2O

Solution:

1) As you examine the unbalanced equation, you see several items that need balancing. Ignoring the water, we have ammonium, sulfate and hydrogen (from the H2C2O4) that need to be balanced. All of these can be balanced with one coefficient:

FeSO4(NH4)2SO4 6H2O + H2C2O4 2H2O ---> FeC2O4 2H2O + 2NH4HSO4 + H2O

2) The only remaining item out of balance is the water. One coefficient change takes care of that:

FeSO4(NH4)2SO4 6H2O + H2C2O4 2H2O ---> FeC2O4 2H2O + 2NH4HSO4 + 6H2O

Problem #68: BCl3 + P4 + H2 ---> BP + HCl

Solution:

1) Balance P:

BCl3 + P4 + H2 ---> 4BP + HCl

2) Balance B:

4BCl3 + P4 + H2 ---> 4BP + HCl

3) Balance Cl:

4BCl3 + P4 + H2 ---> 4BP + 12HCl

4) Balance H:

4BCl3 + P4 + 6H2 ---> 4BP + 12HCl

Problem #69: NH3 + KAl(SO4)2 12H2O ---> Al(OH)3 + (NH4)2SO4 + KOH + H2O

Solution:

NH3 + KAl(SO4)2 12H2O ---> Al(OH)3 + 2(NH4)2SO4 + KOH + H2O

2) Now, balance the nitrogen:

4NH3 + KAl(SO4)2 12H2O ---> Al(OH)3 + 2(NH4)2SO4 + KOH + H2O

3) Balance the hydrogen (look only at the ammonia/ammonium amount of H), the hydroxide and the water all at the same time:

4NH3 + KAl(SO4)2 12H2O ---> Al(OH)3 + 2(NH4)2SO4 + KOH + 8H2O

There are 16 hydrogens in ammonium sulfate. Twelve of them are accounted for in the ammonia.

The four remaining hydrogens come from four waters in the alum. The four hydroxides that result go to the Al(OH)3 and the KOH.

That leaves 8 waters to be balanced and a total of 36 hydrogens on each side.

5) This equation is often seen:

4NH4OH + KAl(SO4)2 12H2O ---> Al(OH)3 + 2(NH4)2SO4 + KOH + 12H2O

Problem #70: FeS2 + O2 ---> Fe2O3 + SO2

1) Balance the iron:

2FeS2 + O2 ---> Fe2O3 + SO2

2) Balance the sulfur:

2FeS2 + O2 ---> Fe2O3 + 4SO2

3) Balance the oxygens using a fraction:

2FeS2 + 112O2 ---> Fe2O3 + 4SO2

4) Multiply through by 2:

4FeS2 + 11O2 ---> 2Fe2O3 + 8SO2

1) Balance the sulfur:

FeS2 + O2 ---> Fe2O3 + 2SO2

2) Balance the iron:

2FeS2 + O2 ---> Fe2O3 + 2SO2

3) Rebalance the sulfur:

2FeS2 + O2 ---> Fe2O3 + 4SO2

4) Balance the oxygen as in solution #1 to arrive at the balanced equation.

This link will take you to the above equation answered on Yahoo Answers. The answerer combines the balancing of the Fe and the S into one step.

Problem #71: C11H12N2O2 + O2 ---> CO2 + H2O + N2

Solution:

1) Balance C:

C11H12N2O2 + O2 ---> 11CO2 + H2O + N2

2) Balance H:

C11H12N2O2 + O2 ---> 11CO2 + 6H2O + N2

3) Since the N is already balanced, the only one left is oxygen:

C11H12N2O2 + 13O2 ---> 11CO2 + 6H2O + N2

Problem #72: FeCl3 6H2O + K2C2O4 H2O ---> K7[Fe(C2O4)5(H2O)4] + KCl + H2O

Solution:

1) Get rid of all the water:

FeCl3 + K2C2O4 ---> K7[Fe(C2O4)5] + KCl

2) We need five oxalates on the left-hand side:

FeCl3 + 5K2C2O4 ---> K7[Fe(C2O4)5] + KCl

3) Balance the potassium with 3KCl on the right:

FeCl3 + 5K2C2O4 ---> K7[Fe(C2O4)5] + 3KCl

This also balances the Cl, the Fe was always in balance.

4) Add the water back in:

FeCl3 6H2O + 5K2C2O4 H2O ---> K7[Fe(C2O4)5(H2O)4] + 3KCl + H2O

5) There are 11 waters on the left-hand side. There are 4 waters on the right-hand side (ignoring the last H2O). Balance the water:

FeCl3 6H2O + 5K2C2O4 H2O ---> K7[Fe(C2O4)5(H2O)4] + 3KCl + 7H2O

Done!

Problem #73: Mg(NO3)2 ---> MgO + NO2 + O2

Solution:

1) The Mg is balanced, the O is in too many places. Balance the N:

Mg(NO3)2 ---> MgO + 2NO2 + O2

2) Count the oxygens to find six on the left and 7 on the right. Reduce the O on the right by 1:

Mg(NO3)2 ---> MgO + 2NO2 + 12O2

3) Clear the fraction:

2Mg(NO3)2 ---> 2MgO + 4NO2 + O2

Problem #74: VF5 + HI ---> V2I10 + HF

Solution:

1) Balance the V:

2VF5 + HI ---> V2I10 + HF

2) Balance the F

2VF5 + HI ---> V2I10 + 10HF

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

2VF5 + 10HI ---> V2I10 + 10HF

Bonus Problem: Na2CO3 + C + N2 ---> NaCN + CO2

Solution:

1) Here's a tempting way to start this problem:

Na2CO3 + 2C + N2 ---> 2NaCN + CO2

Everything is balanced but the oxygens.

2) Balance the oxygens using whole numbers only:

2Na2CO3 + 2C + N2 ---> 2NaCN + 3CO2

This could be a problem for an introductory student to see. That's because it throws stuff already balanced out of balance. Sometimes, that happens.

3) Balance (and re-balance) everything else:

2Na2CO3 + 5C + 2N2 ---> 4NaCN + 3CO2

4) You could use a fraction. Start with the equation from step 1:

Na2CO3 + 2C + N2 ---> 2NaCN + CO2

Balance the oxygen:

Na2CO3 + 52C + N2 ---> 2NaCN + 32CO2

The 52 comes from the 2 already present (think of it as 42) and 12 more C

The problem here is that an introductory student will protest that you cannot have 12 of an atom. That's true, but you can have 12 of a mole. And that is something introductory students, just learning to balance equations, would not necessarily have figured out yet.

Multiply through by 2 to clear the fractions and get the answer given in step 3.

5) Sometimes you can foresee how to deal with the oxygens, since the oxygens only come three at a time on the left and only come two at a time on the right. In which case, you start with the oxygen:

2Na2CO3 + C + N2 ---> NaCN + 3CO2

and proceed to the final answer.

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