# How do you balance K+B_2O_3 -> K_2O + B?

Mar 6, 2016

One way is to use the method of oxidation numbers.

#### Explanation:

$\text{K" + "B"_2"O"_3 → "K"_2"O" + "B}$

Step 1. Identify the atoms that change oxidation number

$\stackrel{\textcolor{b l u e}{0}}{\text{K") + stackrelcolor(blue)(+3)("B")_2stackrelcolor(blue)("-2")("O")_3 → stackrelcolor(blue)(+1)("K")_2stackrelcolor(blue)("-2")("O") + stackrelcolor(blue)(0)("B}}$

The changes in oxidation number are:

$\text{K: 0 → +1; Change =+1 (oxidation)}$
$\text{B: +3 → 0; Change = -3 (reduction)}$

Step 2. Equalize the changes in oxidation number

We need 3 atoms of $\text{K}$ for every 1 atom of $\text{B}$ or 6 atoms of $\text{K}$ for every 2 atoms of $\text{B}$. This gives us total changes of +6 and -6.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{6} \text{K" + color(red)(1)"B"_2"O"_3 → color(red)(3)"K"_2"O" + color(red)(2)"B}$

Every formula now has a coefficient. The equation should be balanced.

Step 4. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l)bb "On the right}}$
$\textcolor{w h i t e}{m m} \text{6 K"color(white)(mmmm) "6 K}$
$\textcolor{w h i t e}{m m} \text{2 B"color(white)(mmmm) "2 B}$
$\textcolor{w h i t e}{m m} \text{3 O"color(white)(mmmm) "3 O}$

The balanced equation is

color(red)(6"K" + "B"_2"O"_3 → 3"K"_2"O" + 2"B"