# How do you balance C_6H_6 + H_2 -> C_6H_12?

Apr 7, 2016

${C}_{6} {H}_{6} + 3 {H}_{2} \to {C}_{6} {H}_{12}$

#### Explanation:

Balancing equations is more of an algorithm than a precise process most of the time. You try one thing, it doesn't work, you try another. It's the way some things go.

To begin with, its helpful to know the amounts of things on each side, how much carbon and hydrogen is reacted and formed afterwards.

On the left we have $6$ carbon and $8$ hydrogen, while on the right we have $6$ carbon and $12$ hydrogen. You want all the numbers for each element to be the same. Carbon is already there, six on both sides, it's just hydrogen that you have to sort out.

You need four more hydrogen to make $8$ into $12$, preferably without changing the number of carbons, though sometimes this is unavoidable and you have to do a few extra steps to sort this new problem, or try something else.

Luckily you have some elemental hydrogen, ${H}_{2}$, which means you can change the number of hydrogen without affecting anything else. When balancing equations, elemental molecules are your friend.

To get $4$ more hydrogen atoms, you need $2$ more ${H}_{2}$ molecules, which makes $3$ in total, rendering the whole equation

${C}_{6} {H}_{6} + 3 {H}_{2} \to {C}_{6} {H}_{12}$

There is $6$ carbon and $12$ hydrogen on the left, and $6$ carbon and $12$ hydrogen on the right. The equation is balanced.