Check all that apply.

ch4(g) + 2o2(g) → co2(g) + 2h2o(g)

-4 for c in ch4

+4 for c in co2

-2 for o in all substances

+1 for h in both ch4 and h2o

+4 for o in h2o

4 for C in CH4

+4 for C in CO2

-2 for O in all substances

+1 for H in both CH4 and H2O

option 1,2,3 and 4 are correct. Option 5 is not correct

Explanation:

Step 1: Data given

Oxidation number of H = +1

Oxidation number of O = -2

Step 2: The balanced equation

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: The oxidation numbers

-4 for C in CH4

 ⇒ Oxidation number of H = +1

    ⇒ 4x +1 = +4

C has an oxidation number of -4

This is correct.

+4 for C in CO2

 ⇒ Oxidation number of O = -2

    ⇒ 2x -2 = -4

C has an oxidation number of +4

This is correct.

-2 for O in all substances

⇒ this is correct, the oxidation number of O is always -2 (except in H2O2 and Na2O2)

This is correct.

+1 for H in both CH4 and H2O

⇒ this is correct, the oxidation number of H is always +1 (except in metal hydrides).

This is correct.

+4 for O in H2O

 ⇒ Oxidation number of H = +1

    ⇒ 2x +1 = +2

The oxidation number of O is -2

This is not correct

it means there are 3 atoms of oxygen