# Question 62a46

May 1, 2017

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

#### Explanation:

$\text{KClO"_3 → "KClO"_4 + "KCl}$

Step 1. Calculate the oxidation numbers of every atom:

stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+5")("Cl")stackrelcolor(blue)("-2")("O")_3 → stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")("Cl")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("Cl")#

Step 2. Identify the atoms that change oxidation number.

The changes in oxidation number are:

$\text{Cl: +5 → +7; Change = +2}$
$\text{Cl: +5 → -1;" color(white)(ll)"Change ="color(white)(m) "-6}$

Step 3. Equalize the changes in oxidation number.

You need $\text{3 atoms of Cl}$ that increase oxidation number for every $\text{1 atom of Cl}$ that decreases oxidation number. . This gives us total changes of $\text{+6}$ and $\text{-6}$.

Step 4. Insert coefficients to get these numbers.

$\text{KClO"_3 → color(red)(3)"KClO"_4 + color(blue)(1)"KCl}$

Step 5. Balance $\text{K}$.

$\textcolor{\mathmr{and} a n \ge}{4} \text{KClO"_3 → color(red)(3)"KClO"_4 + color(blue)(1)"KCl}$

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

Step 6. Check that all atoms balance.

$\text{Atom"color(white)(m) "Left Hand Side"color(white)(m) "Right Hand Side}$
$\textcolor{w h i t e}{m} \text{K} \textcolor{w h i t e}{m m m m m m} 4 \textcolor{w h i t e}{m m m m m m m l} 4$
$\textcolor{w h i t e}{m} \text{Cl} \textcolor{w h i t e}{m m m m m l l} 4 \textcolor{w h i t e}{m m m m m m m l} 4$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m m l} 12 \textcolor{w h i t e}{m m m m m m m} 12$

The balanced equation is

$\text{4KClO"_3 → "3KClO"_4 + "KCl}$

Now, can you use the same technique to balance the second equation?

Here's a video that might help.

May 1, 2017

$14 K C l {O}_{3} + 4 {I}_{2} + 4 H N {O}_{3} + 2 {H}_{2} O \rightarrow 7 K C l {O}_{4} + 7 K C l + 8 H I {O}_{3} + 4 N O$

#### Explanation:

I'm going to assume it is in an acidic solution.

The first step would be to separate the overall equation into two half reactions, but you've already done that. So next would be to balance everything except for the H and O.

$\textcolor{n a v y}{2 K C l {O}_{3} \rightarrow K C l {O}_{4} + K C l}$

$\textcolor{m a r \infty n}{{I}_{2} + H N {O}_{3} \rightarrow 2 H I {O}_{3} + N O + {H}_{2} O}$

Then balance the oxygen by adding H2O.

$\textcolor{n a v y}{2 K C l {O}_{3} \rightarrow K C l {O}_{4} + K C l + 2 {H}_{2} O}$

$\textcolor{m a r \infty n}{{I}_{2} + H N {O}_{3} + 5 {H}_{2} O \rightarrow 2 H I {O}_{3} + N O + {H}_{2} O}$

Now balance the hydrogen by adding H+.

$\textcolor{n a v y}{2 K C l {O}_{3} + 4 {H}^{+} \rightarrow K C l {O}_{4} + K C l + 2 {H}_{2} O}$

$\textcolor{m a r \infty n}{{I}_{2} + H N {O}_{3} + 5 {H}_{2} O \rightarrow 2 H I {O}_{3} + N O + {H}_{2} O + 7 {H}^{+}}$

Add e- as needed to balance the charges on each side of the equation.

$\textcolor{n a v y}{2 K C l {O}_{3} + 4 {H}^{+} + 4 {e}^{-} \rightarrow K C l {O}_{4} + K C l + 2 {H}_{2} O}$

$\textcolor{m a r \infty n}{{I}_{2} + H N {O}_{3} + 5 {H}_{2} O \rightarrow 2 H I {O}_{3} + N O + {H}_{2} O + 7 {H}^{+} + 7 {e}^{-}}$

Finally, multiply each half-reaction by an integer so that when they are added, the e- will cancel out. Then add them and simplify.

7($\textcolor{n a v y}{2 K C l {O}_{3} + 4 {H}^{+} + 4 {e}^{-} \rightarrow K C l {O}_{4} + K C l + 2 {H}_{2} O}$)

+

4($\textcolor{m a r \infty n}{{I}_{2} + H N {O}_{3} + 5 {H}_{2} O \rightarrow 2 H I {O}_{3} + N O + {H}_{2} O + 7 {H}^{+} + 7 {e}^{-}}$)

=

$\textcolor{n a v y}{14 K C l {O}_{3} + 28 {H}^{+} + 28 {e}^{-} + \textcolor{m a r \infty n}{4 {I}_{2} + 4 H N {O}_{3} + 20 {H}_{2} O} \textcolor{b l a c k}{\rightarrow} 7 K C l {O}_{4} + 7 K C l + 14 {H}_{2} O + \textcolor{m a r \infty n}{8 H I {O}_{3} + 4 N O + 4 {H}_{2} O + 28 {H}^{+} + 28 {e}^{-}}}$

$\textcolor{n a v y}{14 K C l {O}_{3} + \cancel{28 {H}^{+}} + \cancel{28 {e}^{-}} + \textcolor{m a r \infty n}{4 {I}_{2} + 4 H N {O}_{3} + \cancel{20} 2 {H}_{2} O} \textcolor{b l a c k}{\rightarrow} 7 K C l {O}_{4} + 7 K C l + \cancel{14 {H}_{2} O} + \textcolor{m a r \infty n}{8 H I {O}_{3} + 4 N O + \cancel{4 {H}_{2} O} + \cancel{28 {H}^{+}} + \cancel{28 {e}^{-}}}}$

$14 K C l {O}_{3} + 4 {I}_{2} + 4 H N {O}_{3} + 2 {H}_{2} O \rightarrow 7 K C l {O}_{4} + 7 K C l + 8 H I {O}_{3} + 4 N O$