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

2 Answers
Nov 12, 2015

Answer:

#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.

Jan 10, 2016

Answer:

#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?