### What if you do not have a net ionic equation?Problems #11 - 25

Problem #11: In the reaction of titanium metal with excess potassium chromate and sulfuric acid, the products are titanium (IV) sulfate, potassium sulfate, chromium (III) sulfate, and water. Write and balance the equation and calculate the coefficient for the potassium sulfate in the balanced equation.

Solution:

1) Write the equation (with correct formulas!):

Ti + K2CrO4 + H2SO4 ---> Ti(SO4)2 + K2SO4 + Cr2(SO4)3 + H2O

2) Convert it to net ionic:

Ti + CrO42- ---> Ti4+ + Cr3+

3) Write and balance half-reactions:

Ti ---> Ti4+
CrO42- ---> Cr3+

Ti ---> Ti4+ + 4e¯
3e¯ + 8H+ + CrO42- ---> Cr3+ + 4H2O

4) Balance net ionic equation:

32H+ + 3Ti + 4CrO42- ---> 3Ti4+ + 4Cr3+ + 16H2O

5) Make chromium(III) sulfate:

20H+ + 6H2SO4 + 3Ti + 4CrO42- ---> 3Ti4+ + 2Cr2(SO4)3 + 16H2O

Note how six sulfuric acid has been added in as well. This balances the six sulfates added.

6) Make titanium(IV) sulfate:

8H+ + 12H2SO4 + 3Ti + 4CrO42- ---> 3Ti(SO4)2 + 2Cr2(SO4)3 + 16H2O

Note how another six sulfuric acid has been added in as well. This balances the six sulfates added. Also, see the hydrogen ion deceasing by 12 each time, for a total of 24 so far.

7) Make potassium chromate:

8H+ + 12H2SO4 + 3Ti + 4K2CrO4 ---> 3Ti(SO4)2 + 2Cr2(SO4)3 + 16H2O + 4K2SO4

Notice how 4 potassium sulfate showed up on the right. That's to balance the potassium. Now, I'll fix the extra 4 sulfates that currently exist on the right-hand side of the equation.

8) Introduce 4 sulfates to the left-hand side of the equation:

16H2SO4 + 3Ti + 4K2CrO4 ---> 3Ti(SO4)2 + 2Cr2(SO4)3 + 16H2O + 4K2SO4

Did you see what I did? I simply introduced 4 more sulfuric acids. I took the 8 hydrogen ion already there and combined them with 4 sulfate ions I needed to add to the left-hand side. Notice that the coefficient of the sulfuric acid went from 12 to 16.

It's balanced and the answer the problem wants is 4.

Problem #12: K2Cr2O7 + SnCl2 + HCl ---> CrCl3 + SnCl4 + KCl + H2O

Solution:

1) Write the net ionic equation:

Cr2O72- + Sn2+ ---> Cr3+ + Sn4+

2) Break into half-reactions and balance:

Cr2O72- ---> Cr3+
Sn2+ ---> Sn4+

6e¯ + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O
Sn2+ ---> Sn4+ + 2e¯

14H+ + 3Sn2+ + Cr2O72- ---> 2Cr3+ + 3Sn4+ + 7H2O

3) Add chloride to the Cr(III) and Sn(IV) ions:

2H+ + 12 HCl + 3SnCl2 + Cr2O72- ---> 2CrCl3 + 3SnCl4 + 7H2O

2H+ + 12HCl + 3SnCl2 + K2Cr2O7 ---> 2CrCl3 + 3SnCl4 + 7H2O + 2KCl

Notice that there are now two extra chlorides on the right-hand side.

5) Add two chlorides to the left-hand side:

14HCl + 3SnCl2 + K2Cr2O7 ---> 2CrCl3 + 3SnCl4 + 7H2O + 2KCl

Problem #13: K2Cr2O7 + FeSO4 + H2SO4 ---> K2SO4 + Fe2(SO4)3 + Cr2(SO4)3 + H2O

Solution:

1) Write the net ionic equation:

Cr2O72¯ + Fe2+ ---> Fe26+ + Cr26+

Notice how I preserved the subscripts for the Fe(III) ion and the Cr(III) ion on the right-hand side. In the next example, I do not employ this little trick. In several other problems, I use the above trick. One way will be effective for some and the other way will be effective for others.

2) Write the half-reactions and balance them:

Cr2O72¯ ---> Cr26+
Fe2+ ---> Fe26+

6e¯ + 14H+ + Cr2O72¯ ---> Cr26+ + 7H2O
2Fe2+ ---> Fe26+ + 2e¯

3) The balanced net ionic equation:

6Fe2+ + 14H+ + Cr2O72¯ ---> Cr26+ + 7H2O + 3Fe26+

4) I'm going to add everything to the left-hand side and nothing to the right-hand side:

6FeSO4 + 7H2SO4 + K2Cr2O7 ---> Cr26+ + 7H2O + 3Fe26+

I added thirteen sulfates and two potassium ions. Now, I have to do the same to the right-hand side.

5) Add sulfate to the chromium(III) ion:

6FeSO4 + 7H2SO4 + K2Cr2O7 ---> Cr2(SO4)3 + 7H2O + 3Fe26+

That's three (of thirteen).

6) Add sulfate to the iron(III) ion:

6FeSO4 + 7H2SO4 + K2Cr2O7 ---> Cr2(SO4)3 + 7H2O + 3Fe2(SO4)3

That's nine more (remember the coefficient of three in front of the iron(III) on the right-hand side), for a total of twelve. One more sulfate to go.

7) Pick up the last sulfate with the two potassium ions still unbalanced:

6FeSO4 + 7H2SO4 + K2Cr2O7 ---> Cr2(SO4)3 + 7H2O + 3Fe2(SO4)3 + K2SO4

Problem #14: C12H22O11 + H2SO4 + K2Cr2O7 ---> CO2 + Cr2(SO4)3 + K2SO4

Solution:

1) Write the net ionic equation:

C12H22O11 + Cr2O72- ---> CO2 + Cr3+

2) Write the half-reactions and balance them:

C12H22O11 ---> CO2
Cr2O72- ---> Cr3+

13H2O + C12H22O11 ---> 12CO2 + 48H+ + 48e¯
6e¯ + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O

3) The balanced net ionic equation:

64H+ + C12H22O11 + 8Cr2O72- ---> 12CO2 + 16Cr3+ + 43H2O
4) Add sulfates to the Cr(III) ion:
16H+ + 24H2SO4 + C12H22O11 + 8Cr2O72- ---> 12CO2 + 8Cr2(SO4)3 + 43H2O

Notice how the sulfates are balanced with sulfuric acid. Notice also that this consumes 48 of the 64 hydrogen ions.

5) Add potassium ion to the dichromate:
16H+ + 24H2SO4 + C12H22O11 + 8K2Cr2O7 ---> 12CO2 + 8Cr2(SO4)3 + 43H2O + 8K2SO4

Notice how the potassium is balanced with potassium sulfate. This introduces eight extra sulfate on the right-hand side.

6) Balance the extra sulfates by making more eight sulfuric acid on the left-hand side:
32H2SO4 + C12H22O11 + 8K2Cr2O7 ---> 12CO2 + 8Cr2(SO4)3 + 43H2O + 8K2SO4

Notice how the 16 remaining hydrogen ions are included in the sulfuric acid.

Problem #15: Na2HAsO3 + KBrO3 + HCl ---> NaCl + KBr + H3AsO4

Solution:

1) Write the net ionic equation:

HAsO32- + BrO3¯ ---> AsO43- + Br¯

2) Write the half-reactions and balance:

HAsO32- ---> AsO43-
BrO3¯ ---> Br¯

H2O + HAsO32- ---> AsO43- + 3H+ + 2e¯
6e¯ + 6H+ + BrO3¯ ---> Br¯ + 3H2O

3HAsO32- + BrO3¯ + 6H+ ---> 3H3AsO4 + Br¯

3) Add one potassium to each side:

3HAsO32- + KBrO3 + 6H+ ---> 3H3AsO4 + KBr

4) Add six sodium and six chloride to the left-hand side:

3Na2HAsO3 + KBrO3 + 6HCl ---> 3H3AsO4 + KBr

5) Add six sodum chloride to the right-hand side:

3Na2HAsO3 + KBrO3 + 6HCl ---> 3H3AsO4 + KBr + 6NaCl

Problem #16: K2Cr2O7 + H2C2O4 · 2H2O ---> K[Cr(C2O4)2(H2O)2] · 2H2O + CO2 + H2O

Solution:

1) Strip out spectator ions:

Cr2O72- + C2O42- ---> Cr3+ + CO2

2) Balance in acidic solution (because of the oxalic acid):

Cr2O72- ---> Cr3+
C2O22- ---> CO2

6e- + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O
C2O42- ---> 2CO2 + 2e-

6e- + 14H+ + Cr2O72- ---> 2Cr3+ + 7H2O
3C2O42- ---> 6CO2 + 6e-

14H+ + Cr2O72- + 3C2O42- ---> 2Cr3+ + 6CO2 + 7H2O

3) To restore the molecular equation the trick, to me, is to restore the right-hand side and then deal with the left:

14H+ + Cr2O72- + 3C2O42- ---> 2K[Cr(C2O4)2(H2O)2] + 6CO2 + 3H2O

Notice how I also reduced the water on the right-hand side. I will also ignore the water of hydration for the moment.

4) The net effect is that I have 4 oxalates and two potassium on the right that still need to be balanced. Le's do both of them:

14H+ + K2Cr2O7 + 7C2O42- ---> 2K[Cr(C2O4)2(H2O)2] + 6CO2 + 3H2O

5) Reform the oxalic acid:

K2Cr2O7 + 7H2C2O4 ---> 2K[Cr(C2O4)2(H2O)2] + 6CO2 + 3H2O

6) When I add in the water of hydration, keep in mind that there is a 7 in front of the oxalic acid. That means 14 waters will be added to the left-hand side. On the right, the waters of hydration will only add 4, so I have to add 10 more to the 3H2O:

K2Cr2O7 + 7H2C2O4 · 2H2O ---> 2K[Cr(C2O4)2(H2O)2] · 2H2O + 6CO2 + 13H2O

7) Check the balance and find this:

42 hydrogens and 49 oxygens; I won't bother with the others.

Problem #17: NaCl + H3PO4 + MnO2 ---> Na3PO4 + Mn3(PO4)2 + H2O + Cl2

Solution:

1) Write the net ionic equation:

Cl¯ + MnO2 ---> Cl2 + Mn2+

The usual practice is keep the formula MnO2 untouched. Eventually, those oxygens will combine with hydrogen ion to make water.

2) Separate into half-reactions and balance:

Cl¯ ---> Cl2
MnO2 ---> Mn2+

2Cl¯ ---> Cl2 + 2e¯
2e¯ + 4H+ + MnO2 ---> Mn2+ + 2H2O

4H+ + 2Cl¯ + MnO2 ---> Mn2+ + Cl2 + 2H2O

4) Notice the formula Mn3(PO4)2. It requires three manganese and we have only one in the net ionic. Multiply through by three:

12H+ + 6Cl¯ + 3MnO2 ---> 3Mn2+ + 3Cl2 + 6H2O

5) I will make the Mn3(PO4)2 by adding in two phosphates:

6H+ + 2H3PO4 + 6Cl¯ + 3MnO2 ---> Mn3(PO4)2 + 3Cl2 + 6H2O

Notice how I also made two phosphoric acids.

6) I'm going to make 2 more phosphoric acid as well as six NaCl:

4H3PO4 + 6NaCl + 3MnO2 ---> Mn3(PO4)2 + 2Na3PO4 + 3Cl2 + 6H2O

Note how two sodium phosphate were added to the right-hand side.

Problem #18: NaCl(aq) + H2SO4(aq) + MnO2(s) ---> Na2SO4(aq) + MnSO4(aq) + H2O(l) + Cl2(g)

Solution:

1) Half-reactions:

MnO2 ---> Mn2+
Cl¯ ---> Cl2

2) Balanced:

2e¯ + 4H+ + MnO2 ---> Mn2+ + 2H2O
2Cl¯ ---> Cl2 + 2e¯

4H+ + MnO2 + 2Cl¯ ---> Mn2+ + 2H2O + Cl2

2H+ + H2SO4 + MnO2 + 2Cl¯ ---> MnSO4 + 2H2O + Cl2

2H+ + H2SO4 + MnO2 + 2NaCl ---> MnSO4 + 2H2O + Cl2 + 2Na+

6) Add in one more sulfate:

2H2SO4 + MnO2 + 2NaCl ---> MnSO4 + 2H2O + Cl2 + Na2SO4

Problem #19: Fe(NO3)2(aq) + K2Cr2O7(aq) + HNO3(aq) ---> Fe(NO3)3(aq) + Cr(NO3)3(aq) + KNO3 + H2O

Solution:

1) Net ionic:

Fe2+ + Cr2O72¯ ---> Fe3+ + Cr3+

2) Half-reactions and balance:

Fe2+ ---> Fe3+
Cr2O72- ---> Cr3+

Fe2+ ---> Fe3+ + e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

6Fe2+ ---> 6Fe3+ + 6e¯
6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O

14H+ + 6Fe2+ + Cr2O72¯ ---> 6Fe3+ + 2Cr3+ + 7H2O

4) Add nitrate (and two potassium ions) to the left-hand side:

14HNO3 + 6Fe(NO3)2 + K2Cr2O7 ---> 6Fe3+ + 2Cr3+ + 7H2O

That's 26 nitrates (and two potassium) that have to go on the right-hand side.

5) Eighteen nitrates go with the 6Fe3+ and six with the 2Cr3+:

14HNO3 + 6Fe(NO3)2 + K2Cr2O7 ---> 6Fe(NO3)3 + 2Cr(NO3)3 + 7H2O

6) That leaves us with two nitrates to account for as well as the two potassium ions:

14HNO3 + 6Fe(NO3)2 + K2Cr2O7 ---> 6Fe(NO3)3 + 2Cr(NO3)3 + 2KNO3 + 7H2O

Problem #20: Na2S2O3 + KBrO3 + H2SO4 ---> Na2S4O6 + K2SO4 + Na2SO4 + Br2 + H2O

Solution:

1) Write balanced net ionic half-reactions:

2S2O32- ---> S4O62- + 2e-
10e- + 12H+ + 2BrO3- ---> Br2 + 6H2O

10S2O32- ---> 5S4O62- + 10e-
10e- + 12H+ + 2BrO3- ---> Br2 + 6H2O

12H+ + 10S2O32- + 2BrO3- ---> 5S4O62- + Br2 + 6H2O

3) Add six sulfates, 20 sodium ions and two potassium ions to the left-hand side only:

6H2SO4 + 10Na2S2O3 + 2KBrO3 --->

4) Only 10 sodium ions can be accounted for by making 5Na25S4O6. Everything else is accounted by in the spectator ions:

10Na2S2O3 + 2KBrO3 + 6H2SO4 ---> 5Na2S4O6 + K2SO4 + 5Na2SO4 + Br2 + 6H2O

Problem #21: K2Cr2O7 + NaCl + H2SO4 ---> Cl2 + H2O + K2SO4 + Na2SO4 + CrSO4

Solution:

1) Write the net ionic:

Cr2O72¯ + Cl¯ ---> Cl2 + Cr2+

2) Write half-reactions and balance:

Cr2O72¯ ---> Cr2+
Cl¯ ---> Cl2

8e¯ + 14H+ + Cr2O72¯ ---> 2Cr2+ + 7H2O
2Cl¯ ---> Cl2 + 2e¯

3) Equalize electrons and write balanced net ionic:

8e¯ + 14H+ + Cr2O72¯ ---> 2Cr2+ + 7H2O
8Cl¯ ---> 4Cl2 + 8e¯

14H+ + Cr2O72¯ + 8Cl¯ ---> 4Cl2 + 2Cr2+ + 7H2O

4) Add everything to the left-hand side:

7H2SO4 + K2Cr2O7 + 8NaCl ---> 4Cl2 + 2Cr2+ + 7H2O

I added two potassium ions, eight sodium ions and seven sulfate ions.

5) Add two potassium ions and one sulfate to the RHS:

7H2SO4 + K2Cr2O7 + 8NaCl ---> 4Cl2 + 2Cr2+ + 7H2O + K2SO4

6) Add eight sodium ions and four sulfate ions to the RHS:

7H2SO4 + K2Cr2O7 + 8NaCl ---> 4Cl2 + 2Cr2+ + 7H2O + K2SO4 + 4Na2SO4

7) Add the last two sulfate ions to create the final answer:

7H2SO4 + K2Cr2O7 + 8NaCl ---> 4Cl2 + 2CrSO4 + 7H2O + K2SO4 + 4Na2SO4

Problem #22: H3AsO4 + Zn + HNO3 ---> AsH3 + Zn(NO3)2 + H2O

1) Write the net ionic half-reactions:

H3AsO4 ---> AsH3
Zn ---> Zn2+

I could have written AsO43¯ if I had wanted to. I would have just re-formed H3AsO4 at the end.

2) Balance in acidic solution:

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

3) Equalize electrons:

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

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

5) Bring back to molecular:

8HNO3 + H3AsO4 + 4Zn ---> AsH3 + 4Zn(NO3)2 + 4H2O

Problem #23: KMnO4 + CH3OH ---> MnO2 + H2O + HCHO + KOH

Solution:

1) Half-reactions:

CH3OH ---> HCHO
MnO4¯ ---> MnO2

2) Balance in acidic solution:

CH3OH ---> HCHO + 2H+ + 2e¯
3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O

3) Equalize electrons:

3CH3OH ---> 3HCHO + 6H+ + 6e¯
6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O

4) Add half reactions and eliminate hydrogen ion and electrons:

2H+ + 2MnO4¯ + 3CH3OH ---> 2MnO2 + 4H2O + 3HCHO

5) Change to basic by adding 2 hydroxides and then eliminating water:

2MnO4¯ + 3CH3OH ---> 2MnO2 + 2H2O + 3HCHO + 2OH¯

6) Add in two potassium ions:

2KMnO4 + 3CH3OH ---> 2MnO2 + 2H2O + 3HCHO + 2KOH

Problem #24: MnSO4 + (NH4)2S2O8 + H2O ----> MnO2 + H2SO4 + (NH4)2SO4

Solution:

1) Half-reactions:

Mn2+ ---> MnO2
S2O82¯ ---> SO42¯

2) Balance:

2H2O + Mn2+ ---> MnO2 + 4H+ + 2e¯
2e¯ + S2O82¯ ---> 2SO42¯

2H2O + Mn2+ + S2O82¯ ---> MnO2 + 2SO42¯ + 4H+

2H2O + MnSO4 + S2O82¯ ---> MnO2 + 3SO42¯ + 4H+

2H2O + MnSO4 + (NH4)2S2O8 ---> MnO2 + (NH4)2SO4 + 2H2SO4

Note that I made two sulfuric acid with the four hydrogen ions and two sulfate ions. The third sulfate ion went to make the ammonium sulfate.

Problem #25: U(SO4)2 + KMnO4 + H2O ---> H2SO4 + K2SO4 + MnSO4 + UO2SO4

Solution:

1) Half-reactions:

U4+ ---> UO22+
MnO4¯ ---> Mn2+

2) Balance:

2H2O + U4+ ---> UO22+ + 4H+ + 2e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

3) Make electrons equal:

10H2O + 5U4+ ---> 5UO22+ + 20H+ + 10e¯
10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O

4) Add and eliminate electrons as well as excess water and hydrogen ion:

2H2O + 5U4+ + 2MnO4¯ ---> 5UO22+ + 2Mn2+ + 4H+

5) Add 10 sulfates and two potassium ions to the left-hand side:

2H2O + 5U(SO4)2 + 2KMnO4 ---> 5UO22+ + 2Mn2+ + 4H+

6) Add five sulfates to UO22+, two sulfates to Mn2+ and two sulfate to the H+:

2H2O + 5U(SO4)2 + 2KMnO4 ---> 5UO2SO4 + 2MnSO4 + 2H2SO4

7) Add one K2SO4 to the right to account for the 10th sulfate and the two potassium ions:

2H2O + 5U(SO4)2 + 2KMnO4 ---> 5UO2SO4 + 2MnSO4 + 2H2SO4 + K2SO4