# Question #6f8ee

Jan 13, 2017

The balanced equation is

${\text{P"_4 + "10H"_2"S""O"_4 → "4H"_3"P""O"_4 + "4H"_2"O" + "10S""O}}_{2}$

#### Explanation:

We can use the method of oxidation numbers to balance this equation.

${\text{P"_4 + "H"_2"S""O"_4 → "H"_3"P""O"_4 + "H"_2"O" + "S""O}}_{2}$

Step 1. Identify the atoms that change oxidation number

${\stackrel{\textcolor{b l u e}{0}}{\text{P")_4 + stackrelcolor(blue)("+1")"H"_2stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 → stackrelcolor(blue)("+1")("H")_3stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") + stackrelcolor(blue)("+4")("S")stackrelcolor(blue)("-2")("O}}}_{2}$

The changes in oxidation number are:

$\text{P:" color(white)(m)0 → "+5"; "Change = +5 (oxidation)}$
$\text{S: +6 → +4; Change ="color(white)(m) "-2 (reduction)}$

Step 2. Equalize the changes in oxidation number

We need 5 atoms of $\text{S}$ for every 2 atoms of $\text{P}$ or 10 atoms of $\text{S}$ for every 4 atoms of $\text{P}$. This gives us total changes of -20 and +20.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{1} {\text{P"_4 + color(red)(10)"H"_2"S""O"_4 → color(red)(4)"H"_3"P""O"_4 + "H"_2"O" + color(red)(10)"S""O}}_{2}$

Step 4. Balance $\text{O}$

We have fixed 40 $\text{O}$ atoms on the left and 36 $\text{O}$ atoms on the right, so we need 4 more $\text{O}$ atoms on the right. Put a 4 before $\text{H"_2"O}$.

$\textcolor{red}{1} {\text{P"_4 + color(red)(10)"H"_2"S""O"_4 → color(red)(4)"H"_3"P""O"_4 + color(blue)(4)"H"_2"O" + color(red)(10)"S""O}}_{2}$

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

Step 7. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l) "On the right}}$
$\textcolor{w h i t e}{m m l l} \text{4 P"color(white)(mmmmm) "4 P}$
$\textcolor{w h i t e}{m m} \text{20 H"color(white)(mmmml) "20 H}$
$\textcolor{w h i t e}{m m} \text{10 S"color(white)(mmmmll) "10 S}$
$\textcolor{w h i t e}{m m} \text{40 O"color(white)(mmmml) "40 O}$

The balanced equation is

$\textcolor{red}{{\text{P"_4 + "10H"_2"S""O"_4 → "4H"_3"P""O"_4 + "4H"_2"O" + "10S""O}}_{2}}$