What if you do not have a net ionic equation?
Problems #26 - 60

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¯

6) Add:

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¯

4) Add:

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¯

4) Add:

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

4) Add:

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⁺

7) Add four more sulfates:

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

4) Add:

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

4) Add:

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 #34a: 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¯

4) Add:

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 #34b: Na₂C₂O₄ + KMnO₄ + H₂SO₄ ---> CO₂ + MnSO₄

Solution:

1) Repeat steps 1-4 of the solution to 34a:

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

2) Add eight sulfates, ten sodium and two potassium to the left-hand side:

8H₂SO₄ + 5Na₂C₂O₄ + 2KMnO₄ ---> 2Mn²⁺ + 10CO₂ + 8H₂O

3) Add two sulfates to the right-hand side:

8H₂SO₄ + 5Na₂C₂O₄ + 2KMnO₄ ---> 2MnSO₄ + 10CO₂ + 8H₂O

4) Add two potassium and one sulfate to the right-hand side:

8H₂SO₄ + 5Na₂C₂O₄ + 2KMnO₄ ---> 2MnSO₄ + 10CO₂ + K₂SO₄ + 8H₂O

5) What remains are ten sodium and five sulfate. Add them in:

8H₂SO₄ + 5Na₂C₂O₄ + 2KMnO₄ ---> 2MnSO₄ + 10CO₂ + K₂SO₄ + 5Na₂SO₄ + 8H₂O

6) I found a solution to 34b here. The person hand wrote the entire solution and photographed each page.

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¯

4) Add:

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 #36a: 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¯

4) Add:

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 more sulfate to the right to balance the sulfates:

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 #36b: H₂SO₄(aq) + KMnO₄(aq) + NaHSO₃(aq) ---> NaHSO₄(aq) + MnSO₄(aq) + K₂SO₄(aq) + H₂O(ℓ)

Solution:

1) Write the net ionic:

MnO₄¯ + HSO₃¯ ---> Mn²⁺ + HSO₄¯

2) Write the half-reactions:

MnO₄¯ ---> Mn²⁺
HSO₃¯ ---> HSO₄¯

3) Balance:

5e¯ + 8H⁺ + MnO₄¯ ---> Mn²⁺ + 4H₂O
H₂O + HSO₃¯ ---> HSO₄¯ + 2H⁺ + 2e¯

4) Equalize electrons:

10e¯ + 16H⁺ + 2MnO₄¯ ---> 2Mn²⁺ + 8H₂O
5H₂O + 5HSO₃¯ ---> 5HSO₄¯ + 10H⁺ + 10e¯

5) Add:

6H⁺ + 2MnO₄¯ + 5HSO₃¯ ---> 2Mn²⁺ + 5HSO₄¯ + 3H₂O

6) Add three sulfate ions, two potassium ions, and five sodium ions to the left-hand side:

3H₂SO₄ + 2KMnO₄ + 5NaHSO₃ ---> 2Mn²⁺ + 5HSO₄¯ + 3H₂O

7) Two sulfates go to the Mn²⁺, the third makes one K₂SO₄ and it's pretty easy to spot where the sodium ions go:

3H₂SO₄ + 2KMnO₄ + 5NaHSO₃ ---> 2MnSO₄ + 5NaHSO₄ + K₂SO₄ + 3H₂O

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¯

4) Add:

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.

Problem #38a: 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¯

4) Add:

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 38b: As₂O₃ + I₂ ---> H₃AsO₄ + HI

Solution:

1) Half-reactions:

As₂O₃ ---> H₃AsO₄
I₂ ---> HI

Note: keep everything molecular.

2) Balance:

5H₂O + As₂O₃ ---> 2H₃AsO₄ + 4H⁺ + 4e¯
2e¯ + 2H⁺ + I₂ ---> 2HI

3) Equalize electrons:

5H₂O + As₂O₃ ---> 2H₃AsO₄ + 4H⁺ + 4e¯
4e¯ + 4H⁺ + 2I₂ ---> 4HI

4) Add:

5H₂O + As₂O₃ + 2I₂ ---> 2H₃AsO₄ + 4HI

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⁺

Add 12 hydroxide:

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

Remove 12 waters:

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

5) Remove factor of 4:

3OH¯ + 3H₂O + P₄ ---> 3H₂PO₂¯ + PH₃

6) Add thee potassium ions:

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¯

4) Add:

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

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¯

4) Add:

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

5) Adding in spectator ions:

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

4) Add:

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¯

4) Add:

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

4) Add:

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¯

4) Add:

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

7) This problem also shows up with two waters of hydration, like this:

FeC₂O₄ ⋅ 2H₂O(s) + H₂O₂(aq) + K₂C₂O₄(aq) ---> K₃[Fe(C₂O₄)₃](aq) + KOH(aq) + H₂O(ℓ)

The balanced equation in that case would be this:

2FeC₂O₄ ⋅ 2H₂O(s) + H₂O₂(aq) + 4K₂C₂O₄(aq) ---> 2K₃[Fe(C₂O₄)₃](aq) + 2KOH(aq) + 4H₂O(ℓ)

The balancing process would be the same as the example above (which lacks the waters of hydration). The water would be added back in as the very last step. The total of four waters of hydration simply become four water molecules in the solution. However, remember to write (ℓ) for the water and not (aq).

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¯

4) Add:

18H⁺ + 3IO₃¯ + 24I¯ ---> 9I₃¯ + 9H₂O

I could take out a factor of three right here, before making everything molecular. I missed that when I wrote the answer, so I decided to leave everything as is. 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

5) Add:

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¯

5) Add:

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

5) Add:

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¯

4) Add:

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

4) Add:

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

5) Convert to basic:

2OH¯ + 2MnO₂ + 3SnO₃²¯ ---> 2MnO₄¯ + 3SnO₂²¯ + H₂O

6) Recover molecular by adding in proper amounts of K and Na:

2KOH + 2MnO₂ + 3Na₂SnO₃ ---> 2KMnO₄ + 3Na₂SnO₂ + H₂O

Problem #53: K₂Cr₂O₇ + Na₂SO₃ + H₂SO₄ ----> Cr₂(SO₄)₃ + K₂SO₄ + Na₂SO₄ + H₂O

Solution:

1) Identify the skeleton net-ionic by removal of all spectator ions:

Cr₂O₇²¯ + SO₃²¯ ---> Cr³⁺ + SO₄²¯

2) Separate into half-reactions:

Cr₂O₇²¯ ---> Cr³⁺
SO₃²¯ ---> SO₄²¯

3) Balance the half-reactions:

6e¯ + 14H⁺ + Cr₂O₇²¯ ---> 2Cr³⁺ + 7H₂O
H₂O + SO₃²¯ ---> SO₄²¯ + 2H⁺ + 2e¯

4) Equalize the electrons:

6e¯ + 14H⁺ + Cr₂O₇²¯ ---> 2Cr³⁺ + 7H₂O
3H₂O + 3SO₃²¯ ---> 3SO₄²¯ + 6H⁺ + 6e¯

5) Add and then eliminate duplicate ions:

8H⁺ + Cr₂O₇²¯ + 3SO₃²¯ ---> Cr₂(SO₄)₃ + 4H₂O

I also reformed the chromium(III) sulfate.

6) Add in a complete set of spectator ions on the left-hand side:

4H₂SO₄ + K₂Cr₂O₇ + 3Na₂SO₃ ---> Cr₂(SO₄)₃ + 4H₂O

I added four sulfate, two potassium, and six sodium.

7) Make one potassium sulfate on the right-hand side:

4H₂SO₄ + K₂Cr₂O₇ + 3Na₂SO₃ ---> Cr₂(SO₄)₃ + K₂SO₄ + 4H₂O

I have three sulfate and six sodium still to balance.

8) Add three sodium sulfate to the right-hand side:

4H₂SO₄ + K₂Cr₂O₇ + 3Na₂SO₃ ---> Cr₂(SO₄)₃ + K₂SO₄ + 3Na₂SO₄ + 4H₂O

Since there is oxygen in every compound, that should serve as a good check for balance. There are 32 oxygens on each side of the arrow. The equation is balanced and we are done.

9) However, you have to be careful. I saw this "balanced" equation on an "answers" website:

7H₂SO₄ + K₂Cr₂O₇ + 12Na₂SO₃ ---> Cr₂(SO₄)₃ + K₂SO₄ + 12Na₂SO₄ + 7H₂O

If you check the oxygens only, you'll find 71 on each side and conclude it's balanced. However, the equation is not balanced. If you check the sulfur, you will find 19 on the left, but only 16 on the right.

Be careful out there on the Interwebz!

Problem #54: Kr₂Cr₂O₇ + C₂H₅OH + H₂SO₄ ---> Cr₂(SO₄)₃ + K₂SO₄ + CH₃COOH + H₂O

Solution:

1) Create a net-ionic equation by removing spectator ions:

Cr₂O₇²¯ + C₂H₅OH ---> Cr³⁺ + CH₃COOH

Note that I kept both ethanol and acetic acid as molecular species.

2) Half-reactions:

Cr₂O₇²¯ ---> Cr³⁺
C₂H₅OH ---> CH₃COOH

3) Balance in acidic solution:

6e¯ + 14H⁺ + Cr₂O₇²¯ ---> 2Cr³⁺ + 7H₂O
H₂O + C₂H₅OH ---> CH₃COOH + 4H⁺ + 4e¯

4) Equalize electrons:

12e¯ + 28H⁺ + 2Cr₂O₇²¯ ---> 4Cr³⁺ + 14H₂O
3H₂O + 3C₂H₅OH ---> 3CH₃COOH + 12H⁺ + 12e¯

5) Add:

16H⁺ + 2Cr₂O₇²¯ + 3C₂H₅OH ---> 4Cr³⁺ + 3CH₃COOH + 11H₂O

6) Restore the molecular equation:

8H₂SO₄ + 2K₂Cr₂O₇ + 3C₂H₅OH ---> 2Cr₂(SO₄)₃ + 3CH₃COOH + 2K₂SO₄ + 11H₂O

I added eight sulfates and four potassium ions to each side.

Problem #55: H₃AsO₄(aq) + H₂O(ℓ) + Zn(s) + KNO₃ ---> AsH₃(g) + KOH(aq) + Zn(NO₃)₂(aq)

Solution:

1) Put into net ionic:

H₃AsO₄ + Zn ---> AsH₃ + Zn²⁺

2) Half-reactions:

H₃AsO₄ ---> AsH₃
Zn ---> Zn²⁺

3) Balance in acidic:

8e¯ + 8H⁺ + H₃AsO₄ ---> AsH₃ + 4H₂O
Zn ---> Zn²⁺ + 2e¯

4) Equalize electrons:

8e¯ + 8H⁺ + H₃AsO₄ ---> AsH₃ + 4H₂O
4Zn ---> 4Zn²⁺ + 8e¯

5) Add:

8H⁺ + H₃AsO₄ + 4Zn ---> AsH₃ + 4Zn²⁺ + 4H₂O

6) Convert to basic:

4H₂O + H₃AsO₄ + 4Zn ---> AsH₃ + 4Zn²⁺ + 8OH¯

Note that four duplicate waters were removed.

7) Convert to molecular by adding eight potassium ions and 8 nitrate ions:

4H₂O + H₃AsO₄ + 4Zn + 8KNO₃ ---> AsH₃ + 4Zn(NO₃)₂ + 8KOH

Problem #56: NaOH + Zn + NaNO₃ ---> Na₂ZnO₂ + NH₃

Solution:

1) Write the net ionic equation:

OH¯ + Zn + NO₃¯ ---> ZnO₂²¯ + NH₃

2) Separate into half-reactions:

Zn ---> ZnO₂²¯
NO₃¯ ---> NH₃

3) Balance as if in acidic solution:

2H₂O + Zn ---> ZnO₂²¯ + 4H⁺ + 2e¯
8e¯ + 9H⁺ + NO₃¯ ---> NH₃ + 3H₂O

4) Equalize electrons:

8H₂O + 4Zn ---> 4ZnO₂²¯ + 16H⁺ + 8e¯
8e¯ + 9H⁺ + NO₃¯ ---> NH₃ + 3H₂O

5) Add:

5H₂O + 4Zn + NO₃¯ ---> 4ZnO₂²¯ + NH₃ + 7H⁺

6) Convert to basic:

7OH¯ + 4Zn + NO₃¯ ---> 4ZnO₂²¯ + NH₃ + 2H₂O

7) Add eight sodium ions to convert to molecular:

7NaOH + 4Zn + NaNO₃ ---> 4Na₂ZnO₂ + NH₃ + 2H₂O

Problem #57: C₂H₅OH + K₂Cr₂O₇ + HCl ---> C₂H₄O₂ + CrCl₃ + KCl + H₂O

Solution:

1) Remove all the spectator ions to get the net ionic:

C₂H₅OH + Cr₂O₇²¯ ---> C₂H₄O₂ + Cr³⁺

2) Separate into half-reactions:

C₂H₅OH ---> C₂H₄O₂
Cr₂O₇²¯ ---> Cr³⁺

3) Balance in acidic solution:

H₂O + C₂H₅OH ---> C₂H₄O₂ + 4H⁺ + 4e¯
6e¯ + 14H⁺ + Cr₂O₇²¯ ---> 2Cr³⁺ + 7H₂O

4) Equalize electrons:

3H₂O + 3C₂H₅OH ---> 3C₂H₄O₂ + 12H⁺ + 12e¯
12e¯ + 28H⁺ + 2Cr₂O₇²¯ ---> 4Cr³⁺ + 14H₂O

5) Add:

16H⁺ + 3C₂H₅OH + 2Cr₂O₇²¯ ---> 3C₂H₄O₂ + 4Cr³⁺ + 11H₂O

6) Add 16 chlorides and 4 potassium ions on the left-hand side:

16HCl + 3C₂H₅OH + 2K₂Cr₂O₇ ---> 3C₂H₄O₂ + 4Cr³⁺ + 11H₂O

7) Add 12 chlorides to the right-hand side:

16HCl + 3C₂H₅OH + 2K₂Cr₂O₇ ---> 3C₂H₄O₂ + 4CrCl₃ + 11H₂O

8) Add four KCl to the right-hand side to finish:

16HCl + 3C₂H₅OH + 2K₂Cr₂O₇ ---> 3C₂H₄O₂ + 4CrCl₃ + 4KCl + 11H₂O

Problem #58: HCl + K₂CrO₄ + FeCl₂ ---> FeCl₃ + CrCl₃ + KCl + H₂O

Solution:

1) Write the net-ionic equation:

CrO₄²¯ + Fe²⁺ ---> Fe³⁺ + Cr³⁺

2) Half-reactions:

CrO₄²¯ ---> Cr³⁺
Fe²⁺ ---> Fe³⁺

3) Balance in acidic solution:

3e¯ + 8H⁺ + CrO₄²¯ ---> Cr³⁺ + 4H₂O
Fe²⁺ ---> Fe³⁺ + e¯

3e¯ + 8H⁺ + CrO₄²¯ ---> Cr³⁺ + 4H₂O
3Fe²⁺ ---> 3Fe³⁺ + 3e¯

5) Add:

8H⁺ + CrO₄²¯ + 3Fe²⁺ ---> 3Fe³⁺ + Cr³⁺ + 4H₂O

6) Make molecular oon the left-hand side only:

8HCl + K₂CrO₄ + 3FeCl₂ ---> 3Fe³⁺ + Cr³⁺ + 4H₂O

14 chlorides and two potassium ions were added.

7) Add chlorides to the right-hand side:

8HCl + K₂CrO₄ + 3FeCl₂ ---> 3FeCl₃ + CrCl₃ + 4H₂O

8) 12 chlorides were added. Add two KCl to RHS to finish the balancing:

8HCl + K₂CrO₄ + 3FeCl₂ ---> 3FeCl₃ + CrCl₃ + 2KCl + 4H₂O

Problem #59: Bi(NO₃)₃ ⋅ 5H₂O + H₂O₂ + RuCl₃ + NaOH ---> Bi₂Ru₂O₇ + NaNO₃ + NaCl + H₂O

Solution:

1) A bit of discussion:

The name of Bi₂Ru₂O₇ is bismuth ruthenate(IV). That Roman numeral means the oxidation state of the Ru is +4. Two of them means +8 and, when coupled with the −14 of the O, we have −6 left to be offset by two Bi. That makes each Bi a +3. The bismuth is neither oxidized or reduced in this reaction.

2) This is the net ionic:

H₂O₂ + Ru³⁺ ---> Ru₂O₇⁶¯ + H₂O

3) Half-reactions:

H₂O₂ ---> H₂O
Ru³⁺ ---> Ru₂O₇⁶¯

2e¯ + 2H⁺ + H₂O₂ ---> 2H₂O
7H₂O + 2Ru³⁺ ---> Ru₂O₇⁶¯ + 14H⁺ + 2e¯

5) Add:

5H₂O + 2Ru³⁺ + H₂O₂ ---> Ru₂O₇⁶¯ + 12H⁺

6) Change to basic:

12OH¯ + 2Ru³⁺ + H₂O₂ ---> Ru₂O₇⁶¯ + 7H₂O

7) Add in 12 sodium ion and six chloride ions:

12NaOH + 2RuCl₃ + H₂O₂ ---> Ru₂O₇⁶¯ + 7H₂O + 6NaCl + 6Na⁺

8) Add in two bismuth(III) nitrate pentahydrate:

2Bi(NO₃)₃ ⋅ 5H₂O + 12NaOH + 2RuCl₃ + H₂O₂ ---> Bi₂Ru₂O₇ + 17H₂O + 6NaCl + 6NaNO₃

Note: when you add in two Bi(NO₃)₃ ⋅ 5H₂O, you are adding 10 water, not 5. That's why the water on the right is 17 and not 12.

Problem #60: MnSO₄ + NaBiO₃ + H₂SO₄ ---> NaMnO₄ + Bi₂(SO₄)₃ + H₂O + Na₂SO₄

Solution:

1) Write the net-ionic equation:

Mn²⁺ + BiO₃¯ ---> MnO₄¯ + Bi³⁺

2) Half-reactions:

Mn²⁺ ---> MnO₄¯
BiO₃¯ ---> Bi³⁺

3) Balance:

4H₂O + Mn²⁺ ---> MnO₄¯ + 8H⁺ + 5e¯
2e¯ + 6H⁺ + BiO₃¯ ---> Bi³⁺ + 3H₂O

4) Equalize electrons:

8H₂O + 2Mn²⁺ ---> 2MnO₄¯ + 16H⁺ + 10e¯
10e¯ + 30H⁺ + 5BiO₃¯ ---> 5Bi³⁺ + 15H₂O

5) Add:

14H⁺ + 2Mn²⁺ + 5BiO₃¯ ---> 2MnO₄¯ + 5Bi³⁺ + 7H₂O

6) An interesting problem arises. The bismuth sulfate uses up Bi two at a time and we have five Bi present. Using the Bi two at a time leaves us with an extra Bi on the right. The solution is to double the balanced net-ionic reaction.

28H⁺ + 4Mn²⁺ + 10BiO₃¯ ---> 4MnO₄¯ + 10Bi³⁺ + 14H₂O

7) Add 18 sulfates and 10 sodium ions to the left side:

14H₂SO₄ + 4MnSO₄ + 10NaBiO₃ ---> 4MnO₄¯ + 10Bi³⁺ + 14H₂O

8) Fifteen sulfates go to the bismuth(III) sulfate and 4 sodium ions to the sodium permanganate:

14H₂SO₄ + 4MnSO₄ + 10NaBiO₃ ---> 4NaMnO₄ + 5Bi₂(SO₄)₃ + 14H₂O

9) That leaves six sodium ions and three sulfates still unaccounted for. We will use them to make sodium sulfate:

14H₂SO₄ + 4MnSO₄ + 10NaBiO₃ ---> 4NaMnO₄ + 5Bi₂(SO₄)₃ + 3Na₂SO₄ + 14H₂O