# How do you balance this chemical equation? "KMnO"_4 + "Na"_2"C"_2"O"_4 + "H"_2"SO"_4 → "K"_2"SO"_4 + "Na"_2"SO"_4 + "MnSO"_4 + "CO"_2 + "H"_2"O"

Aug 24, 2014

$\textcolor{b l u e}{\text{2KMnO"_4 + "5Na"_2"C"_2"O"_4 + "8H"_2"SO"_4 → "K"_2"SO"_4 + "5Na"_2"SO"_4 + "MnSO"_4 + "10CO"_2 + "8H"_2"O}}$

#### Explanation:

One method is to balance equations by the oxidation number method.

Step 1. Calculate the oxidation numbers of every atom:

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{K")stackrelcolor(blue)(+7)("Mn")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)(+1)("Na")_2stackrelcolor(blue)(+3)("C")_2stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)(+1)("H")_2stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 → stackrelcolor(blue)(+1)("K")_2stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)(+1)("Na")_2stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)(+2)("Mn")stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)(+4)("C")stackrelcolor(blue)("-2")("O")_2 + stackrelcolor(blue)(+1)("H")_2stackrelcolor(blue)("-2")("O}}$

Step 2. Identify the atoms that change oxidation number.

$\text{Mn: +7 → +2; Change = -5}$
$\text{C: +3 → +4; Change = +1}$

Step 3. Equalize the changes in oxidation number.

You need $\text{1 atom of Mn}$ for every $\text{5 atoms of C}$ or $\text{2 atoms of Mn}$ for every $\text{10 atoms of C}$. This gives us total changes of $+ 10$ and $\text{-10}$.

Step 4. Insert coefficients to get these numbers.

$\textcolor{red}{2} \text{KMnO"_4 + color(red)(5)"Na"_2"C"_2"O"_4 + "H"_2"SO"_4 → "K"_2"SO"_4 + "Na"_2"SO"_4 + color(red)(2)"MnSO"_4 + color(red)(10)"CO"_2 + "H"_2"O}$

Step 5 Balance $\text{K}$.

$\textcolor{red}{2} \text{KMnO"_4 + color(red)(5)"Na"_2"C"_2"O"_4 + "H"_2"SO"_4 → color(blue)(1)"K"_2"SO"_4 + "Na"_2"SO"_4 + color(red)(2)"MnSO"_4 + color(red)(10)"CO"_2 + "H"_2"O}$

Step 6 Balance $\text{Na}$.

$\textcolor{red}{2} \text{KMnO"_4 + color(red)(5)"Na"_2"C"_2"O"_4 + "H"_2"SO"_4 → color(blue)(1)"K"_2"SO"_4 + color(orange)(5)"Na"_2"SO"_4 + color(red)(2)"MnSO"_4 + color(red)(10)"CO"_2 + "H"_2"O}$

Step 7. Balance $\text{S}$.

$\textcolor{red}{2} \text{KMnO"_4 + color(red)(5)"Na"_2"C"_2"O"_4 + color(olive)(8)"H"_2"SO"_4 → color(blue)(1)"K"_2"SO"_4 + color(orange)(5)"Na"_2"SO"_4 + color(red)(2)"MnSO"_4 + color(red)(10)"CO"_2 + "H"_2"O}$

Step 8. Balance $\text{O}$.

$\textcolor{red}{2} \text{KMnO"_4 + color(red)(5)"Na"_2"C"_2"O"_4 + color(olive)(8)"H"_2"SO"_4 → color(blue)(1)"K"_2"SO"_4 + color(orange)(5)"Na"_2"SO"_4 + color(red)(2)"MnSO"_4 + color(red)(10)"CO"_2 + color(brown)(8)"H"_2"O}$

Every substance now has a coefficient, so the equation should be balanced.

Step 9. Check that all atoms balance.

$\boldsymbol{\text{Atom"color(white)(m) bb"Left Hand Side"color(white)(m) bb"Right Hand Side}}$
$\textcolor{w h i t e}{m} \text{K} \textcolor{w h i t e}{m m m m m} 2 \textcolor{w h i t e}{m m m m m m m m m m} 2$
$\textcolor{w h i t e}{m} \text{Mn} \textcolor{w h i t e}{m m m m l} 2 \textcolor{w h i t e}{m m m m m m m m m m} 2$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m l l} 60 \textcolor{w h i t e}{m m m m m m m m m l} 60$
$\textcolor{w h i t e}{m} \text{Na} \textcolor{w h i t e}{m m m m} 10 \textcolor{w h i t e}{m m m m m m m m m l} 10$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{m m m m l l} 10 \textcolor{w h i t e}{m m m m m m m m m l} 10$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m l l} 16 \textcolor{w h i t e}{m m m m m m m m m l} 16$
$\textcolor{w h i t e}{m} \text{S} \textcolor{w h i t e}{m m m m m l} 8 \textcolor{w h i t e}{m m m m m m m m m m} 8$

The balanced equation is

$\text{2KMnO"_4 + "5Na"_2"C"_2"O"_4 + "8H"_2"SO"_4 → "K"_2"SO"_4 + "5Na"_2"SO"_4 + "MnSO"_4 + "10CO"_2 + "8H"_2"O}$

Here's a video on balancing equations by the oxidation number method.