I need to balance this equation by redox reactions with half reactions method

NH4ClO4 + Al --> Al2O3 + AlCl3 + NO

IT wont balance ever, by any method. You have H on the left, not on the right.

Bob is wrong, by half reactions the hydrogen will balance out with the addition of H2O on the right hand side... i'm doing this problem myself and having trouble separating into half reactions in the first place

To balance the given equation using the half-reaction method for redox reactions, follow these steps:

Step 1: Split the equation into two half-reactions. One for the oxidation and one for the reduction.

Oxidation half-reaction:
NH4ClO4 → NO

Reduction half-reaction:
Al → Al2O3 + AlCl3

Step 2: Balance the atoms (other than oxygen and hydrogen) in each half-reaction.

For the oxidation half-reaction:
There are already an equal number of atoms on both sides, so no further balancing is needed.

For the reduction half-reaction:
Balance Al by adding a coefficient of 2 in front of Al:
2 Al → Al2O3 + AlCl3

Step 3: Balance the oxygen atoms by adding water (H2O) molecules.

For the oxidation half-reaction:
Add 4 H2O molecules on the right side to balance the number of oxygen atoms:
NH4ClO4 → NO + 4 H2O

For the reduction half-reaction:
There are already an equal number of oxygen atoms on both sides, so no further balancing is needed.

Step 4: Balance the hydrogen atoms by adding H+ ions.

For the oxidation half-reaction:
Add 8 H+ ions on the left side to balance the number of hydrogen atoms:
NH4ClO4 + 8 H+ → NO + 4 H2O

For the reduction half-reaction:
Add 6 H+ ions on the left side to balance the number of hydrogen atoms:
2 Al + 6 H+ → Al2O3 + AlCl3

Step 5: Balance the charges by adding electrons (e-).

For the oxidation half-reaction:
Add 8 electrons (8e-) on the left side to balance the charges:
NH4ClO4 + 8 H+ + 8e- → NO + 4 H2O

For the reduction half-reaction:
No additional electrons are needed because the charge is already balanced.

Step 6: Multiply the half-reactions by appropriate coefficients to equalize the number of electrons in both half-reactions.

To equalize the number of electrons, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 8:

3(NH4ClO4 + 8 H+ + 8e-) → 3(NO + 4 H2O)
8(2 Al + 6 H+ → Al2O3 + AlCl3)

Simplified balanced equation:
3 NH4ClO4 + 18 H+ + 18e- → 3 NO + 12 H2O
16 Al + 48 H+ → 8 Al2O3 + 8 AlCl3

Step 7: Cancel out the electrons (e-) that appear on both sides of the equation.

The final balanced equation by redox reactions with half-reactions method is:
3 NH4ClO4 + 16 Al → 3 NO + 8 Al2O3 + 8 AlCl3