Question

Can you write and balance the equation for the complete combustion of ethane, `C_2H_6`?

Answer 1

`2C_2H_6+7O_2` `->` `4O_2+6H_2O`

Explanation:

In a combustion reaction with a hydrocarbon in the reactant side you will always have `O_2` as another reactant. As you will always have `CO_2` and `H_2O` as the products.

Knowing that much you can set up your reaction equation..

`C_2H_6 + O_2` `->` `CO_2 + H_2O`

Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.

Start with the `C` atoms first and move to the `H` atoms next. It's easier to leave the `O_2` to the last, it has a way to alter the equation.

Initially, you would arrive at this, before the `O_2` has been balanced:

`C_2H_6 + O_2 -> 2CO_2 + 3H_2O`

But, as you can see, you have an odd amount of `O_2` on the product side. In this case, you have to find the common factor of the amount of `O` on the product side and 2, Because of the `O_2` diatom. Therefore, 14 would be the lowest common factor of 2 and 7.

Answer 2

`C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(g)`

Explanation:

Any hydrocarbon combusts completely to give carbon dioxide and water. Is it balanced? The oxygen bears a half-coefficient. How could I remove it?

Can you represent the combustion of propane and butane with similar reactions?