Help with determining oxidation numbers

Give oxidation numbers for each element in the reactions below. Also, if you specify which substances are being oxidized or reduced, I'd really appreciate it!

1. Al(s) + O2(g) ---> Al2O3(s)

2. NO3-(aq) + S(s) ---> SO2(g) + NO(g)


Thank you!

Ben doesn't appear to understand that these are redox reactions and the oxidation numbers change. The oxidation number is written above the element.
....0.........0.............+3..-2
1. Al(s) + O2(g) ---> Al2O3(s)

...+5-2............0.........+4-2.....…
2. NO3-(aq) + S(s) ---> SO2(g) + NO(g)

1. Al to Al3+ Elements alone are zero. Al has lost 3 electrons, so it is oxidized.
O2 to 2)2- 2 Oxygens have gained 4 electrons. Cross multiply to even out the loss and gain of electrons
4Al + 3O2 = 2Al2O3.
NO31- each oxygen is 2- x3=6- for the overal charge to be 1- the N must be 5+
S is zero and goes to 4+, which balances with the 2 oxygens.The sulphur goes from zero to 4+
4NO3- + 3S + 4H+ = 3SO2 + 4NO + 2H2O
An element that loses electrons is oxidized, and the element that gains electrons is reduced. (More negative.)

1. Al(s) + O2(g) ---> Al2O3(s)

In Reactants
Aluminum = 0
Oxygen = 0

In Products
Aluminum = 3+
Oxygen = 2-

Oxygen is being reduced and Aluminum is being oxidized

2. NO3-(aq) + S(s) ---> SO2(g) + NO(g)

In Reactants
Nitrogen = 5+
Oxygen = 2-
Sulfur = 0

In Products
Nitrogen = 2+
Oxygen = 2-
Sulfur = 4+

Sulfur gets oxidized. Nitrogen gets reduced.

1. Al(s) + O2(g) ---> Al2O3(s)
0 0 +3 -2

Al is oxidized, O2 is reduced
2. NO3-(aq) + S(s) ---> SO2(g) + NO(g)
+5-2 0 +4 -2 +2 -2
N is reduced S is oxidized

Al 3+, O 2-
S 4+, N 2+