# How do you balance K+MgBr -> KBr + Mg?

Apr 6, 2017

First it necessary to balance all of the compounds. Then make sure that there is the same number of atoms on each side of the equation.

#### Explanation:

K(s) = 0 charge.
Mg = +2 charge.
Br = Br-1 charge.
Mg(s) = 0 charge.

$M g B r = M {g}^{+} 2 B {r}_{2}^{-} 1$( balanced compound)
$K B r = {K}^{+} 1 B {r}^{-} 1$ ( balanced compound)

 K(s_ + MgBr_2 = KBr + Mg(s) There are two Br in the reactants but only 1 Br in the products so the KBr must be multiplied by 2

$K \left(s\right) + M g B {r}_{2} = 2 K B r + M g \left(s\right)$ Now there are two K in the product but only 1 K in the reactants so the K must be multiplied by 2

$2 K \left(s\right) + M g B {r}_{2} = 2 K B r + M g \left(s\right)$

This is the balanced chemical reaction
2K in the reactants 2 K in the products.
1 Mg in the reactants 1 Mg in the products
2 Br in the reactants 2 Br in the products.

This is an example of a single replacement reaction.

Apr 6, 2017

Refer to the explanation.
The balanced equation is $\text{2K" + "MgBr"_2}$$\rightarrow$$\text{2KBr" + "Mg}$.

#### Explanation:

First you need to make sure the compounds have the correct formulas. $\text{MgBr}$ is not the correct formula for magnesium bromide. The correct formula is $\text{MgBr"_2}$, because the magnesium ion has a charge of ${2}^{+}$ and each bromide ion has a charge of ${1}^{-}$. So two bromide ions are needed to bond with magnesium in order to make the compound neutral.

Now, we need to balance the equation. A balanced equation has the same number of atoms of each element on both sides. We do that by changing the amount of the reactants and products as needed by adding coefficients (numbers in front) of the formulas. Each coefficient is multiplied by the subscripts for each element. We never change a formula, so never try to balance by changing subscripts. No coefficient or subscript is understood to be $1$.

color(purple)("K" + "MgBr"_2"$\rightarrow$color(purple)("KBr" + "Mg"

There are two bromide ions on the left side and one on the right. So we need to place a coefficient of $2$ in front of $\text{KBr}$

$\textcolor{p u r p \le}{\text{K" + "MgBr"_2}}$$\rightarrow$$\textcolor{m a \ge n t a}{2} \textcolor{p u r p \le}{\text{KBr" + "Mg}}$

Now we have two potassium ions on the right side, but only one on the left side. So we need to place a coefficient of $2$ in front of $\text{K}$.

$\textcolor{m a \ge n t a}{2} \textcolor{p u r p \le}{\text{K" + "MgBr"_2}}$$\rightarrow$$\textcolor{m a \ge n t a}{2} \textcolor{p u r p \le}{\text{KBr" + "Mg}}$

Now we need to verify that the equation has the same number of atoms of each element on each side.

Left:
$\text{2 K}$
$\text{1 Mg}$
$\text{2 Br}$

Right:
$\text{2 K}$
$\text{1 Mg}$
$\text{2 Br}$

So both sides have the same number of atoms of each element, and the equation is balanced.