When ethane is burnt in excess of oxygen, the oxidation number of carbon changes byA. +8B. +7C. +3D. +4

When ethane (C2H6) is burned in excess oxygen, it undergoes complete combustion to produce carbon dioxide (CO2) and water (H2O). Let's calculate the oxidation number change of carbon in this reaction.

The oxidation state of hydrogen (H) is usually +1, and the sum of the oxidation states in a compound is equal to the overall charge of the compound, which is 0 for a neutral molecule.

In carbon dioxide (CO2), the oxidation state of oxygen (O) is typically -2. Since there are two oxygen atoms in CO2, the total oxidation state contributed by oxygen is -2 * 2 = -4.

Now, we can set up an equation to calculate the oxidation state of carbon (C):

2C + (-4) = 0

Solving for 2C:

2C = 4

C = 2

So, the oxidation state of carbon in carbon dioxide (CO2) is +2.

Now, let's consider the oxidation state of carbon in ethane (C2H6). Since hydrogen has an oxidation state of +1 and there are 6 hydrogen atoms, the total oxidation state contributed by hydrogen is +1 * 6 = +6.

Now, we can set up an equation to calculate the oxidation state of carbon in ethane:

2C + (+6) = 0

Solving for 2C:

2C = -6

C = -3

So, the oxidation state of carbon in ethane (C2H6) is -3.

Now, let's find the change in the oxidation number of carbon:

Change = Final oxidation state - Initial oxidation state
Change = (+2) - (-3)
Change = +2 + 3
Change = +5

The oxidation number of carbon changes by +5 during the combustion of ethane in excess oxygen. None of the provided answer choices (A, B, C, D) matches this value.