# How would you balance Br2 + Kl = KBr + I2?

Jun 28, 2018

By considering that atoms must be conserved in chemical reactions, we can figure out that it must be $1 B {r}_{2} + 2 K I = 2 K B r + {I}_{2}$ with a bit of logic.

#### Explanation:

The main concept that must be applied to determine the coefficients (amount of each item) is that there must be equal amounts of each element on each side of the equation. We are not destroying or creating new atoms. In this case, the unbalanced reaction formula is:

$B {r}_{2} + K I = K B r + {I}_{2}$

There are a two problems we need to solve before it will be balanced:

1. There are two moles of Iodine atoms ($I$) on the right side of the equation, while there is only one mole on the right side.

2. There are two moles of bromine ($B r$) atoms on the left side, while there is only one on the right.

Since there are two moles of bromine atoms on the left side, we need two moles on the right as well. We can do this by adding a coefficient of two to the $' K B r '$ term in the equation. Our now modified equation looks like this:

$B {r}_{2} + K I = 2 K B r + {I}_{2}$

There is one mole of Iodine atoms on the left, and two on the right. To fix this, we add a coefficient of two to the $' K I '$ term. The resulting equation is below.

$B {r}_{2} + 2 K I = 2 K B r + {I}_{2}$

Bonus step: We can also put ones in front of the coefficient-less species. This is like changing a phrase from "an apple" to "1 apple". It is the exact same thing, but makes it a little more clear sometimes. This would like like this:

$1 B {r}_{2} + 2 K I = 2 K B r + 1 {I}_{2}$

Can you see that there is now an equal amount of each element on each side of the equation? That means that it is balanced.

Generally if you have basic chemistry questions I'd recommend you check out something like Khan Academy first where they have tons of video tutorials on this sort of thing. Just search "khan academy balancing chemical equations" in google, and you'll find a bunch of awesome videos. Anyway, I'm not sure if you read all of this, because it's a very long answer for a short question, but if you did, it was my pleasure! :)