Question

Write balanced net ionic equations for the following reactions in acid solution.

a. Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed.

b. Solid phosphorus (P4) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate (H2PO4 -) ions.

c. Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese (II) ions are formed.

Answer 1

3N2H4(l) + 2NaBrO3 --> 3N2 + 2NaBr + 6H2O
H in N2H4 = -1, N in N2H2 = +2...H in H2O = +1, N in N2 = 0
Br in NaBrO3 = +5...Br in NaBr = -1

H -1 --> +1 oxidized
N +2 --> 0 reduced
Br +5 --> -1 reduced

1. balance elements oxidized/reduced
4H- --> 2H2O
2N+2 --> N2
BrO3- --> Br-
2. balance the oxygens by adding water to the opposite side.
4H- + 2H2O --> 2H2O
2N+2 --> N2
BrO3- --> Br- + 3H2O
3. balance hydrogens by adding H+ to the opposite side
4H- + 2H2O --> 2H2O + 4H+
2N+2 --> N2
BrO3- + 6H+ --> Br- + 3H2O
4. balance total charge
4H- + 2H2O --> 2H2O + 4H+ + 8e-
2N+2 + 2e- --> N2
BrO3- + 6H+ + 6e- --> Br- + 3H2O
5. combine all 1/2 reactions canceling out elements/compound that are the same on both sides
4H- + 2H2O + 2N+2 + 2e- + BrO3- + 6H+ + 6e- --> 2H2O + 2H+ + 8e- + N2 + Br- + 3H2O
4H- + 2N+2 + BrO3- + 2H+ --> 3H2O + N2 + Br-

2. P4 + 8HNO3 --> 8NO + 4H2PO4-
P in P4 = 0, P in H2PO4- = +5 oxidized
N in HNO3 = +5... N in NO = +1 reduced
1. balance elements being oxidized/reduced
P4 --> 4PO4-3
NO3- --> NO
2. balance oxygens by adding water to the opposite side
P4 + 16H2O --> 4PO4-3
NO3- --> NO + 2H2O
3. balance hydrogens by adding H+ to the opposite side
P4 +16H2O --> 4PO4-3 + 32H+
NO3- + 4H+ --> NO + 2H2O
4. balance charges
P4 + 16H2O --> 4PO4-3 + 32H+ + 20e-
NO3- + 4H+ + 3e- --> NO + 2H2O
5. make the electrons equal in both equations
3P4 + 48H2O --> 12PO4-3 + 96H+ + 60e-
20NO3- + 80H+ + 60e- --> 20NO + 40H2O
6. combine and eliminate
3P4 + 48H2O + 20NO3- + 80H+ + 60e- --> 12PO4-3 + 96H+ + 60e- + 20NO + 40H2O
3P4 + 8H2O + 20NO3- --> 12PO4-3 + 16H+ + 20NO

3. 3 K₂SO₃ + 2 KMnO₄ + H₂O = 2 MnO₂ + 3 K₂SO₄ + 2 KOH

3K2SO3 + 2KMnO4 + H2O --> 2Mn2+ + 2O2- + 4K+ + 3SO4-2 + 2K+ + 2OH-