# How do you balance Mn + HI -> H_2 + MnI_3?

Mar 5, 2016

You can balance it by inspection.

#### Explanation:

The procedure I usually use is:

2. Balance all atoms other than $\text{H}$ and $\text{O}$.
3. Balance $\text{O}$.
4. Balance $\text{H}$.
5. If necessary, multiply by an integer to remove fractions.
6. Check that all atoms are balanced.

${\text{Mn"color(white)(l) + "HI" → "H"_2 + "MnI}}_{3}$

Put a 1 in front of ${\text{MnI}}_{3}$. This number is fixed and does not change until (if necessary) Step 5.

${\text{Mn"color(white)(l) + "HI" → "H"_2 + color(red)(1)"MnI}}_{3}$

Step 2. Balance $\text{Mn}$.

We have fixed $\text{1 Mn}$ on the right, so we need $\text{1 Mn}$ on the left.

$\textcolor{b l u e}{1} {\text{Mn"+ "HI" → "H"_2 + color(red)(1)"MnI}}_{3}$

Step 3. Balance $\text{I}$.

We have fixed $\text{3 I}$ on the right, so we need $\text{3 I}$ on the left.

Put a 3 before $\text{HI}$.

$\textcolor{b l u e}{1} {\text{Mn"+ color(orange)(3)"HI" → "H"_2 + color(red)(1)"MnI}}_{3}$

Step 4. Balance $\text{H}$.

We have fixed $\text{3 H}$ on the left, so we need $\text{3 H}$ on the right.

We have $\text{2 H}$ atoms on the right, so we must multiply by 1.5 to get $\text{3 H}$ atoms.

$\textcolor{b l u e}{1} {\text{Mn"+ color(orange)(3)"HI" → color(green)(1.5)"H"_2 + color(red)(1)"MnI}}_{3}$

All atoms should now between balanced, but we have a fraction in front of ${\text{H}}_{2}$

Step 5. Multiply all coefficients by 2 to remove fractions.

$2 {\text{Mn"+ 6"HI" → 3"H"_2 + "MnI}}_{3}$

Step 6. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l) bb"On the right}}$
$\textcolor{w h i t e}{m l l} \text{2 Mn"color(white)(mmmm) "2 Mn}$
$\textcolor{w h i t e}{m l l} \text{6 H"color(white)(mmmmm) "6 H}$
$\textcolor{w h i t e}{m l l} \text{6 I"color(white)(mmmmml) "6 I}$

The balanced equation is

$\textcolor{red}{2 {\text{Mn"+ 6"HI" → 3"H"_2 + "MnI}}_{3}}$