### What if you do not have a net ionic equation?Problems #26 - 50

Problem #26: H2C2O4(aq) + K2CrO4(aq) + HCl(aq) ---> CrCl3(aq) + KCl(aq) + H2O(ℓ) + CO2(g)

Solution:

1) The half-reactions are these:

CrO42¯ ---> Cr3+
C2O42¯ ---> CO2

2) Balance for oxygen (C also gets balanced):

CrO42¯ ---> Cr3+ + 4H2O
C2O42¯ ---> 2CO2

3) Balance for H:

8H+ + CrO42¯ ---> Cr3+ + 4H2O
C2O42¯ ---> 2CO2

4) Balance for charge:

3e¯ + 8H+ + CrO42¯ ---> Cr3+ + 4H2O
C2O42¯ ---> 2CO2 + 2e¯

5) Equalize charge:

6e¯ + 16H+ + 2CrO42¯ ---> 2Cr3+ + 8H2O
3C2O42¯ ---> 6CO2 + 6e¯

16H+ + 3C2O42¯ + 2CrO42¯ ---> 2Cr3+ + 6CO2 + 8H2O

7) To recover the molecular, you can add stuff back in to the left-hand side:

10HCl + 3H2C2O4 + 2K2CrO4 ---> 2Cr3+ + 6CO2 + 8H2O

I added 10 Cl¯ (note six H+ went to the H2C2O4) and I also added four K+

8) Now, we need to add the same amount to the right-hand side:

10HCl + 3H2C2O4 + 2K2CrO4 ---> 2CrCl3 + 4KCl + 6CO2 + 8H2O

Six Cl¯ went to make the 2CrCl3 and the other four Cl¯ paired with the four K+ to make 4KCl.

Problem #27: Mn(NO3)2 + PbO2 + HNO3 ---> HMnO4 + Pb(NO3)2 + H2O

Solution:

1) Half-reactions:

Mn2+ ---> MnO4¯
PbO2 ---> Pb2+

2) Balance:

4H2O + Mn2+ ---> MnO4¯ + 8H+ + 5e¯
2e¯ + 4H+ + PbO2 ---> Pb2+ + 2H2O

3) Equalize electrons:

8H2O + 2Mn2+ ---> 2MnO4¯ + 16H+ + 10e¯
10e¯ + 20H+ + 5PbO2 ---> 5Pb2+ + 10H2O

4) Add. Eliminate electrons as well as excess hydrogen ions and water:

4H+ + 2Mn2+ + 5PbO2 ---> 5Pb2+ + 2MnO4¯ + 2H2O

5) What needs to be added in is mostly nitrate. The only other item would be two hydrogens intended to make HMnO4. Add them in before adding the nitrates:

6H+ + 2Mn2+ + 5PbO2 ---> 5Pb2+ + 2HMnO4 + 2H2O

I made the decision to add hydrogen back in first based mainly on experience. It now makes it possible to add in the necessary nitrates in one step.

6) Ten nitrates finnish the problem:

6HNO3 + 2Mn(NO3)2 + 5PbO2 ---> 5Pb(NO3)2 + 2HMnO4 + 2H2O

You are welcome to add some nitrate in first and then add in two hydrogens. What will happen is that you'll have two unattached nitrates on the left that will become HNO3 when you add in the two hydrogen.

Problem #28: Sb2O5 + Kl + HCl ---> SbCl3 + KCl + I2 + H2O

Solution:

1) Half-reactions:

Sb2O5 ---> Sb3+
I¯ ---> I2

2) Balance:

4e¯ + 10H+ + Sb2O5 ---> 2Sb3+ + 5H2O
2I¯ ---> I2 + 2e¯

3) Equalize electrons:

4e¯ + 10H+ + Sb2O5 ---> 2Sb3+ + 5H2O
4I¯ ---> 2I2 + 4e¯

10H+ + Sb2O5 + 4I¯ ---> 2Sb3+ + 2I2 + 5H2O

5) Add 10 chlorides to each side:

10HCl + Sb2O5 + 4I¯ ---> 2SbCl3 + 4Cl¯ + 2I2 + 5H2O

Note that 4 chlorides on the right are left as ions.

6) Add four potassium ions to each side:

10HCl + Sb2O5 + 4KI ---> 2SbCl3 + 4KCl + 2I2 + 5H2O

Problem #29: Bi(OH)3 + K2SnO2 ---> Bi + K2SnO3 + H2O

Solution:

1) Half-reactions:

Bi(OH)3 ---> Bi
SnO22¯ ---> SnO32¯

2) Balance:

3e¯ + Bi(OH)3 ---> Bi + 3OH¯
H2O + SnO22¯ ---> SnO32¯ + 2H+ + 2e¯

It does not matter that one half-reaction is in basic and the other in acid. That will be fixed.

3) Equalize electrons:

6e¯ + 2Bi(OH)3 ---> 2Bi + 6OH¯
3H2O + 3SnO22¯ ---> 3SnO32¯ + 6H+ + 6e¯

3H2O + 2Bi(OH)3 + 3SnO22¯ ---> 2Bi + 3SnO32¯ + 6H+ + 6OH¯

6H+ + 6OH¯ = 6H2O and then cancel three H2O.

2Bi(OH)3 + 3SnO22¯ ---> 2Bi + 3SnO32¯ + 3H2O

5) Add six potassium ions to each side:

2Bi(OH)3 + 3K2SnO2 ---> 2Bi + 3K2SnO3 + 3H2O

Problem #30: C14H12 + Na2Cr2O7 + H2SO4 ---> C14H8O2 + Cr2(SO4)3

Solution:

1) Half-reactions:

C14H12 ---> C14H8O2
Cr2O72¯ ---> Cr3+

2) Balance:

2H2O + C14H12 ---> C14H8O2 + 8H+ + 8e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

3) Equalize electrons:

6H2O + 3C14H12 ---> 3C14H8O2 + 24H+ + 24e¯
24e¯ + 56H+ + 4Cr2O72¯ ---> 8Cr3+ + 28H2O

32H+ + 3C14H12 + 4Cr2O72¯ ---> 3C14H8O2 + 8Cr3+ + 22H2O

5) Add 12 sulfates to both sides (to make the Cr2(SO4)3):

8H+ + 12H2SO4 + 3C14H12 + 4Cr2O72¯ ---> 3C14H8O2 + 4Cr2(SO4)3 + 22H2O

6) Add eight sodium ions to each side (to make the Na2Cr2O7):

8H+ + 12H2SO4 + 3C14H12 + 4Na2Cr2O7 ---> 3C14H8O2 + 4Cr2(SO4)3 + 22H2O + 8Na+

16H2SO4 + 3C14H12 + 4Na2Cr2O7 ---> 3C14H8O2 + 4Cr2(SO4)3 + 22H2O + 4Na2SO4

This last step made four more sulfuric acids as well as four sodium sulfates. Note how the sodium sulfate was not part of the equation that was originally given.

Problem #31: C6H6O2 + Na2Cr2O7 + H2SO4 ---> C6H4O2 + Cr2(SO4)3

Solution:

1) Half-reactions:

C6H6O2 ---> C6H4O2
Cr2O72¯ ---> Cr3+

2) Balance:

C6H6O2 ---> C6H4O2 + 2H+ + 2e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

3) Equalize electrons:

3C6H6O2 ---> 3C6H4O2 + 6H+ + 6e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

8H+ + Cr2O72¯ + 3C6H6O2 ---> 3C6H4O2 + 2Cr3+ + 7H2O

5) Put three sulfates in (to make Cr2(SO4)3):

2H+ + 3H2SO4 + Cr2O72¯ + 3C6H6O2 ---> 3C6H4O2 + Cr2(SO4)3 + 7H2O

6) Add two sodium ions and one sulfate to the left:

4H2SO4 + Na2Cr2O7 + 3C6H6O2 ---> 3C6H4O2 + Cr2(SO4)3 + 7H2O + Na2SO4

I also added one sodium sulfate to the right-hand side to balance everything.

Problem #32: C6H10O + HNO3 ---> C6H10O4 + NO2

Solution:

1) Half-reactions:

C6H10O ---> C6H10O4
NO3¯ ---> NO2

2) Balance:

3H2O + C6H10O ---> C6H10O4 + 6H+ + 6e¯
e¯ + 2H+ + NO3¯ ---> NO2 + H2O

3) Equalize electrons:

3H2O + C6H10O ---> C6H10O4 + 6H+ + 6e¯
6e¯ + 12H+ + 6NO3¯ ---> 6NO2 + 6H2O

6H+ + 6NO3¯ + C6H10O ---> 6NO2 + C6H10O4 + 3H2O

5) Nothing needs to be added back in. Only one recombination:

6HNO3 + C6H10O ---> 6NO2 + C6H10O4 + 3H2O

Problem #33: KMnO4 + Na2SO3 + NaOH ---> K2MnO4 + Na2MnO4 + Na2SO4 + H2O

Solution:

1) Half-reactions:

MnO4¯ ---> MnO42¯
SO32¯ ---> SO42¯

2) Balance:

e¯ + MnO4¯ ---> MnO42¯
H2O + SO32¯ ---> SO42¯ + 2H+ + 2e¯

3) Equalize electrons:

2e¯ + 2MnO4¯ ---> 2MnO42¯
H2O + SO32¯ ---> SO42¯ + 2H+ + 2e¯

4) Add and convert to basic:

H2O + 2MnO4¯ + SO32¯ ---> 2MnO42¯ + SO42¯ + 2H+

2OH¯ + 2MnO4¯ + SO32¯ ---> 2MnO42¯ + SO42¯ + H2O

5) Add four sodium ion and two potassium ion to each side:

First, add to the left-hand side:

2NaOH + 2KMnO4 + Na2SO3 ---> 2MnO42¯ + SO42¯ + H2O

Now, to the RHS:

2NaOH + 2KMnO4 + Na2SO3 ---> K2MnO4 + Na2MnO4 + Na2SO4 + H2O

In case you were mildly curious, depending on the compounds selected by the question writer, the following answers could have been written:

2NaOH + 2KMnO4 + Na2SO3 ---> 2Na2MnO4 + K2SO4 + H2O

or

2NaOH + 2NaMnO4 + Na2SO3 ---> 2Na2MnO4 + Na2SO4 + H2O

Problem #34: KMnO4 + H2C2O4 + H2SO4 ---> K2SO4 + MnSO4 + CO2 + H2O

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
C2O42¯ ---> CO2

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
C2O42¯ ---> 2CO2 + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5C2O42¯ ---> 10CO2 + 10e¯

16H+ + 5C2O42¯ + 2MnO4¯ ---> 2Mn2+ + 10CO2 + 8H2O

5) Make five undissociated oxalic acids:

6H+ + 5H2C2O4 + 2MnO4¯ ---> 2Mn2+ + 10CO2 + 8H2O

6) Add two sulfates to each side:

2H+ + 2H2SO4 + 5H2C2O4 + 2MnO4¯ ---> 2MnSO4 + 10CO2 + 8H2O

7) Add two potassium ions and one more sulfate:

3H2SO4 + 5H2C2O4 + 2KMnO4 ---> 2MnSO4 + 10CO2 + K2SO4 + 8H2O

8) The same problem with nitric acid:

KMnO4 + H2C2O4 + HNO3 ---> CO2 + Mn(NO3)2 + KNO3 + H2O

balanced:

2KMnO4 + 5H2C2O4 + 6HNO3 ---> 10CO2 + 2Mn(NO3)2 + 2KNO3 + 8H2O

Problem #35: KMnO4 + H2SO3 ---> K2SO4 + MnSO4 + H2SO4 + H2O

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
SO32¯ ---> SO42¯

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O + SO32¯ ---> SO42¯ + 2H+ + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5H2O + 5SO32¯ ---> 5SO42¯ + 10H+ + 10e¯

6H+ + 2MnO4¯ + 5SO32¯ ---> 2Mn2+ + 5SO42¯ + 3H2O

5) Add four hydrogen ion to each side:

2MnO4¯ + 5H2SO3 ---> 2MnSO4 + 2H2SO4 + SO42¯ + 3H2O

Two sulfates went to sulfuric acid, two sulfates went to MnSO4 and one sulfate is still to be dealt with.

6) Add two potassium ions to finish it:

2KMnO4 + 5H2SO3 ---> 2MnSO4 + 2H2SO4 + K2SO4 + 3H2O

Problem #36: KMnO4(aq) + NaHSO3(aq) + H2SO4(aq) ---> MnSO4(aq) + K2SO4(aq) + Na2SO4(aq) + H2O(ℓ)

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
HSO3¯ ---> SO42¯

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O + HSO3¯ ---> SO42¯ + 3H+ + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5H2O + 5HSO3¯ ---> 5SO42¯ + 15H+ + 10e¯

H+ + 2MnO4¯ + 5HSO3¯ ---> 2Mn2+ + 5SO42¯ + 3H2O

5) We know that there will be at least one H2SO4, but we have only one H+ on the left-hand side. Double everything to get two H+:

2H+ + 4MnO4¯ + 10HSO3¯ ---> 4Mn2+ + 10SO42¯ + 6H2O

6) Add one sulfate, four potassium ions and ten sodium ions to the left-hand side:

H2SO4 + 4KMnO4 + 10NaHSO3 ---> 4Mn2+ + 10SO42¯ + 6H2O

7) I'm going to reunite the 4Mn2+ with four sulfates and I will add in one sulfate to the right:

H2SO4 + 4KMnO4 + 10NaHSO3 ---> 4MnSO4 + 7SO42¯ + 6H2O

8) Add in the 10 sodium ions (using up five sulfates) and the four potassium (using up two sulfates):

H2SO4 + 4KMnO4 + 10NaHSO3 ---> 4MnSO4 + 2K2SO4 + 5Na2SO4 + 6H2O

Problem #37: KMnO4 + Na2SO3 + H2SO4 ---> K2SO4 + MnSO4 + Na2SO4 + H2O

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
SO32¯ ---> SO42¯

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O + SO32¯ ---> SO42¯ + 2H+ + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5H2O + 5SO32¯ ---> 5SO42¯ + 10H+ + 10e¯

6H+ + 2MnO4¯ + 5SO32¯ ---> 2Mn2+ + 5SO42¯ + 3H2O

5) Add all the required spectator ions, but only do it on the left-hand side:

3H2SO4 + 2KMnO4 + 5Na2SO3 ---> 2Mn2+ + 5SO42¯ + 3H2O

I added 3 sulfate ions, two potassium ions, and 10 sodium ions.

6) Add 10 sodium ions to the right-hand side:

3H2SO4 + 2KMnO4 + 5Na2SO3 ---> 2Mn2+ + 5Na2SO4 + 3H2O

I did that simply because I had five sulfate ions already present to take up the ten sodium ions.

7) Now, the three sulfate ions:

3H2SO4 + 2KMnO4 + 5Na2SO3 ---> K2SO4 + 2MnSO4 + 5Na2SO4 + 3H2O

Two of the sulfates went to the potassium sulfate and the other one went to the manganese(II) sulfate. Notice I also added in the two potassium ions.

And it's done.

Problem #38: As2O3 + Cl2 + H2O ---> H3AsO4 + HCl

Solution:

1) Half-reactions:

As2O3 ---> AsO43¯
Cl2 ---> Cl¯

2) Balance:

5H2O + As2O3 ---> 2AsO43¯ + 10H+ + 4e¯
2e¯ + Cl2 ---> 2Cl¯

3) Equalize electrons:

5H2O + As2O3 ---> 2AsO43¯ + 10H+ + 4e¯
4e¯ + 2Cl2 ---> 4Cl¯

5H2O + As2O3 + 2Cl2 ---> 2AsO43¯ + 4Cl¯ + 10H+

5) Reunite ions on the right-hand side:

5H2O + As2O3 + 2Cl2 ---> 2H3AsO4 + 4HCl

Note that no spectator ions needed to be added in.

Problem #39: H2O(ℓ) + P4(s) + KOH(aq) ---> KH2PO2(aq) + PH3(g)

Solution:

1) Half-reactions:

P4 ---> H2PO2¯
P4 ---> PH3

2) Balance in acid (change to basic at the end):

8H2O + P4 ---> 4H2PO2¯ + 8H+ + 4e¯
12e¯ + 12H+ + P4 ---> 4PH3

3) Equalize electrons (still in acid):

24H2O + 3P4 ---> 12H2PO2¯ + 24H+ + 12e¯
12e¯ + 12H+ + P4 ---> 4PH3

4) Add (then change to basic):

24H2O + 4P4 ---> 12H2PO2¯ + 4PH3 + 12H+

12OH¯ + 24H2O + 4P4 ---> 12H2PO2¯ + 4PH3 + 12H2O

Remove 12 waters:

12OH¯ + 12H2O + 4P4 ---> 12H2PO2¯ + 4PH3

5) Remove factor of 4:

3OH¯ + 3H2O + P4 ---> 3H2PO2¯ + PH3

3KOH + 3H2O + P4 ---> 3KH2PO2 + PH3

Problem #40: K2CrO4 + Na2SO3 + HCl --> KCl + Na2SO4 + CrCl3 + H2O

Solution:

1) Half-reactions:

CrO42¯ ---> Cr3+
SO32¯ ---> SO42¯

2) Balance:

3e¯ + 8H+ + CrO42¯ ---> Cr3+ + 4H2O
H2O + SO32¯ ---> SO42¯ + 2H+ + 2e¯

3) Equalize electrons:

6e¯ + 16H+ + 2CrO42¯ ---> 2Cr3+ + 8H2O
3H2O + 3SO32¯ ---> 3SO42¯ + 6H+ + 6e¯

10H+ + 2CrO42¯ + 3SO32¯ ---> 2Cr3+ + 3SO42¯ + 5H2O

5) Add 10 chlorides, four potassium ions and six sodium ions to the left-hand side:

10HCl + 2K2CrO4 + 3Na2SO3 ---> 2Cr3+ + 3SO42¯ + 5H2O

6) On the right-hand side:

six chlorides go to the 2Cr3+
six sodium go to the 3SO42¯
The remaining four K+ and four Cl¯ for 4KCl

10HCl + 2K2CrO4 + 3Na2SO3 ---> 2CrCl3 + 3Na2SO4 + 4KCl + 5H2O

Problem #41: Na2S2O3 + KMnO4 + H2O ---> Na2SO4 + K2SO4 + MnO2 + KOH

Solution:

1) Half-reactions:

S2O32¯ ---> SO42¯
MnO4¯ ---> MnO2

2) Balance (in acid, change to base later):

5H2O + S2O32¯ ---> 2SO42¯ + 10H+ + 8e¯
3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
3) Equalize electrons:
15H2O + 3S2O32¯ ---> 6SO42¯ + 30H+ + 24e¯
24e¯ + 32H+ + 8MnO4¯ ---> 8MnO2 + 16H2O

4) Add, then change to basic:

2H+ + 3S2O32¯ + 8MnO4¯ ---> 8MnO2 + 6SO42¯ + H2O

Add two hydroxide to each side:

H2O + 3S2O32¯ + 8MnO4¯ ---> 8MnO2 + 6SO42¯ + 2OH¯

One excess water was also eliminated.

5) Add six sodium and eight potassium to the left-hand side:

H2O + 3Na2S2O3; + 8KMnO4 ---> 8MnO2 + 6SO42¯ + 2OH¯

6) On the right-hand side, the six sodium will be distributed among three sulfates. Six potassium will take care of the other three sulfates and the last two potassium will form KOH.

H2O + 3Na2S2O3 + 8KMnO4 ---> 8MnO2 + 3Na2SO4 + 3K2SO4 + 2KOH

Problem #42: H2SO4 + KMnO4 + H2O2 ---> MnSO4 + O2 + H2O

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
H2O2 ---> O2

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O2 ---> O2 + 2H+ + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5H2O2 ---> 5O2 + 10H+ + 10e¯

6H+ + 2MnO4¯ + 5H2O2 ---> 2Mn2+ + 5O2 + 8H2O

3H2SO4 + 2KMnO4 + 5H2O2 ---> 2MnSO4 + 5O2 + 8H2O + K2SO4

Three sulfates and two potassium ions were added back in.

Problem #43: K2S + KMnO4 ---> S8 + MnO2 + KOH

Solution:

1) Half-reactions:

S2¯ ---> S8
MnO4¯ ---> MnO2

2) Balance as if in acidic solution:

8S2¯ ---> S8 + 16e¯
3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O

3) Equalize electrons:

24S2¯ ---> 3S8 + 48e¯
48e¯ + 64H+ + 16MnO4¯ ---> 16MnO2 + 32H2O

64H+ + 24S2¯ + 16MnO4¯ ---> 3S8 + 16MnO2 + 32H2O

5) Change to basic by adding 64 hydroxide ions to each side:

64H2O + 24S2¯ + 16MnO4¯ ---> 3S8 + 16MnO2 + 32H2O + 64OH¯

eliminate the water:

32H2O + 24S2¯ + 16MnO4¯ ---> 3S8 + 16MnO2 + 64OH¯

6) The only spectator ion is potassium:

32H2O + 24K2S + 16KMnO4 ---> 3S8 + 16MnO2 + 64KOH

Problem #44: HNO3(aq) + H2S(aq) ---> NO(g) + S8(s) + H2O(ℓ)

Solution:

1) Half-reactions:

NO3¯ ---> NO
S2¯ ---> S8

2) Balance:

3e¯ + 4H+ + NO3¯ ---> NO + 2H2O
8S2¯ ---> S8 + 16e¯

3) Equalize electrons:

48e¯ + 64H+ + 16NO3¯ ---> 16NO + 32H2O
24S2¯ ---> 3S8 + 48e¯

64H+ + 16NO3¯ + 24S2¯ ---> 16NO + 3S8 + 32H2O

5) Distribute the 64 hydrogen ions:

16HNO3 + 24H2S ---> 16NO + 3S8 + 32H2O

Problem #45: Na2S2O3 + KMnO4 + HNO3 ---> MnSO4 + Na2SO4 + NaNO3 + KNO3 + H2O

Solution:

1) Half-reactions:

S2O32¯ ---> SO42¯
MnO4¯ ---> Mn2+

2) Balance:

5H2O + S2O32¯ ---> 2SO42¯ + 10H+ + 8e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

3) Equalize electrons:

25H2O + 5S2O32¯ ---> 10SO42¯ + 50H+ + 40e¯
40e¯ + 64H+ + 8MnO4¯ ---> 8Mn2+ + 32H2O

14H+ + 5S2O32¯ + 8MnO4¯ ---> 8Mn2+ + 10SO42¯ + 7H2O

5) Add 14 nitrates, 10 sodium and 8 potassium to the left:

14HNO3 + 5Na2S2O3 + 8KMnO4 ---> 8MnSO4 + 2SO42¯ + 7H2O

I also formed 8 MnSO4 on the right side.

6) Four sodium with the two sulfates, the other six sodium with six nitrates and the eight potassium with the remaining 8 nitrates:

14HNO3 + 5Na2S2O3 + 8KMnO4 ---> 8MnSO4 + 2Na2SO4 + 6NaNO3 + 8KNO3 + 7H2O

Problem #46: FeC2O4(s) + H2O2(aq) + K2C2O4(aq) ---> K3[Fe(C2O4)3](aq) + KOH(aq)

Solution:

1) Half-reactions:

Fe2+ ---> Fe3+
H2O2 ---> OH¯

2) Balance:

Fe2+ ---> Fe3+ + e¯
2e¯ + H2O2 ---> 2OH¯ <--- no need to balance in acid, then shift to base

3) Equalize electrons:

2Fe2+ ---> 2Fe3+ + 2e¯
2e¯ + H2O2 ---> 2OH¯

2Fe2+ + H2O2 ---> 2Fe3+ + 2OH¯

5) Make molecular formulas on the right:

2Fe2+ + H2O2 ---> 2K3[Fe(C2O4)3] + 2KOH

I added eight potassium ions as well as six oxalate ions.

6) Add two oxalates to the two ferrous ions and make four potassium oxalates:

2FeC2O4 + H2O2 + 4K2C2O4 ---> 2K3[Fe(C2O4)3] + 2KOH

Problem #47: KIO3 + KI + H2SO4 ---> KI3 + K2SO4 + H2O

Solution:

1) Half-reactions:

IO3¯ ---> I3¯
I¯ ---> I3¯

2) Balance:

16e¯ + 18H+ + 3IO3¯ ---> I3¯ + 9H2O
3I¯ ---> I3¯ + 2e¯

3) Equalize electrons:

16e¯ + 18H+ + 3IO3¯ ---> I3¯ + 9H2O
24I¯ ---> 8I3¯ + 16e¯

18H+ + 3IO3¯ + 24I¯ ---> 9I3¯ + 9H2O <--- I could take out a factor of three before making everything molecular. I missed that when I answered this question on Yahoo Answers, but I get rid of the factor of 3 at the end.

5) Make the left-hand side molecular:

9H2SO4 + 3KIO3 + 24KI ---> 9I3¯ + 9H2O

I did that by adding nine sulfates and and 27 potassium ion.

6) Add 9 potassium ion to the right:

9H2SO4 + 3KIO3 + 24KI ---> 9KI3 + 9H2O

7) I have 18 K+ and 9 SO42¯ still to add:

9H2OSO4 + 3KIO3 + 24KI ---> 9KI3 + 9K2SO4 + 9H2O

8) Remove a factor of 3:

3H2SO4 + KIO3 + 8KI ---> 3KI3 + 3K2SO4 + 3H2O

Problem #48: C2H5OH + H2CrO4 + H2SO4 ---> CH3CHO + Cr2(SO4)3 + H2O

Solution:

1) Write the net-ionic equation:

C2H5OH + CrO42¯ ---> CH3CHO + Cr3+

2) Half-reactions:

C2H5OH ---> CH3CHO
CrO42¯ ---> Cr3+

3) Balance in acidic solution:

C2H5OH ---> CH3CHO + 2H+ + 2e¯
3e¯ + 8H+ + CrO42¯ ---> Cr3+ + 4H2O

4) Equalize electrons:

3C2H5OH ---> 3CH3CHO + 6H+ + 6e¯
6e¯ + 16H+ + 2CrO42¯ ---> 2Cr3+ + 8H2O

3C2H5OH + 2CrO42¯ + 10H+ ---> 3CH3CHO + 2Cr3+ + 8H2O

6) Add in three sulfate ions:

3C2H5OH + 2H2CrO4 + 3H2SO4 ---> 3CH3CHO + Cr2(SO4)3 + 8H2O

Problem #49: K2Cr2O7 + HCl ---> CrCl3 + KCl + H2O + Cl2

Solution:

1) Write the net-ionic equation:

Cr2O72¯ + Cl¯ ---> Cr3+ + Cl2

2) Half-reactions:

Cr2O72¯ ---> Cr3+
Cl¯ ---> Cl2

3) Balance half-reactions in acidic solution:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
2Cl¯ ---> Cl2 + 2e¯

4) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6Cl¯ ---> 3Cl2 + 6e¯

14H+ + Cr2O72¯ + 6Cl¯ ---> 2Cr3+ + 3Cl2 + 7H2O

6) Restore molecular equation:

14H+ + K2Cr2O7 + 6Cl¯ ---> 2Cr3+ + 3Cl2 + 2K+ + 7H2O <--- add two potassium ion

14H+ + K2Cr2O7 + 14Cl¯ ---> 2CrCl3 + 3Cl2 + 2KCl + 7H2O <--- add eight chloride to each side

K2Cr2O7 + 14HCl ---> 2CrCl3 + 3Cl2 + 2KCl + 7H2O <--- form 14Hcl on the left-hand side

Problem #50: As2O3 + HNO3 ---> H3AsO4 + NO2 + H2O

Solution:

1) Write the net-ionic equation:

As2O3 + NO3¯ ---> H3AsO4 + NO2

2) Half-reactions:

As2O3 ---> H3AsO4
NO3¯ ---> NO2

3) Balance:

5H2O + As2O3 ---> 2H3AsO4 + 4H+ + 4e¯
e¯ + 2H+ + NO3¯ ---> NO2 + H2O

4) Equalize electrons:

5H2O + As2O3 ---> 2H3AsO4 + 4H+ + 4e¯
4e¯ + 8H+ + 4NO3¯ ---> 4NO2 + 4H2O

H2O + As2O3 + 4HNO3 ---> 2H3AsO4 + 4NO2

I put the 4 hydrogen ions and the 4 nitrate ions back together as well.

Problem #51: Cl2 + AgNO3 + H2O ---> AgCl + HClO3 + HNO3

Solution:

1) Net-ionic half-reactions:

Cl2 ---> Cl¯
Cl2 ---> ClO3¯

I'll put the silver ion back in when I reconstitute the molecular equation.

2) Balance:

2e¯ + Cl2 ---> 2Cl¯
6H2O + Cl2 ---> 2ClO3¯ + 12H+ + 10e¯

3) Equalize electrons:

10e¯ + 5Cl2 ---> 10Cl¯
6H2O + Cl2 ---> 2ClO3¯ + 12H+ + 10e¯

6H2O + 6Cl2 ---> 10Cl¯ + 2ClO3¯ + 12H+

5) I need to add in 10 silver atoms. I'll add it in on the left as 10AgNO3

6H2O + 10AgNO3 + 6Cl2 ---> 10Cl¯ + 2ClO3¯ + 12H+

6) I will make 10AgCl on the right and I'll leave the 10 nitrates unbalanced until the next step:

6H2O + 10AgNO3 + 6Cl2 ---> 10AgCl + 2ClO3¯ + 12H+

7) The 10 nitrates still to be balanced will make 10HNO3:

6H2O + 10AgNO3 + 6Cl2 ---> 10AgCl + 2HClO3 + 10HNO3

I used the other two hydrogen ions on the right to make 2HClO3

Problem #52: MnO2 + Na2SnO3 + KOH ---> KMnO4 + Na2SnO2 + H2O

Solution:

1) Half-reactions (note that I took it to net-ionic at the same time):

MnO2 ---> MnO4¯
SnO32¯ ---> SnO22¯

2) Balance in acidic solution (will convert to basic in step 5):

2H2O + MnO2 ---> MnO4¯ + 4H+ + 3e¯
2e¯ + 2H+ + SnO32¯ ---> SnO22¯ + H2O

3) Equalize electrons:

4H2O + 2MnO2 ---> 2MnO4¯ + 8H+ + 6e¯
6e¯ + 6H+ + 3SnO32¯ ---> 3SnO22¯ + 3H2O