# Question 64b84

Apr 13, 2017

$\text{2K"_2"Cr"_2"O"_7 + "C"_2"H"_5"OH" + "16HCl" → "4CrCl"_3 + "11H"_2"O" + "2CO"_2 + "4KCl}$

#### Explanation:

One way is to use the oxidation number method.

$\text{K"_2"Cr"_2"O"_7 + "C"_2"H"_5"OH" + "HCl" → "CrCl"_3 + "H"_2"O" + "CO"_2 + "KCl}$

Step 1. Identify the atoms that change oxidation number

stackrelcolor(blue)("+1")("K")_2stackrelcolor(blue)("+6")("Cr")_2stackrelcolor(blue)("-2")("O")_7 + stackrelcolor(blue)("-2")("C")_2stackrelcolor(blue)("+1")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-1")("Cl") → stackrelcolor(blue)("+3")("Cr")stackrelcolor(blue)("-1")("Cl")_3 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-2")("O")_2 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("Cl")#

The atoms that change oxidation number are:

$\text{Cr: +6 → +3; Change ="color(white)(lll) "-3 (reduction)}$
$\text{C:"color(white)(ml) "-2 → +4; Change = +6 (oxidation)}$

Step 2. Equalize the changes in oxidation number

We need 2 atoms of $\text{Cr}$ for every 1 atom of $\text{C}$ or 4 atoms of $\text{Cr}$ for every 2 atoms
of $\text{C}$.

This gives us total changes of -12 and +12.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{2} \text{K"_2"Cr"_2"O"_7 + color(red)(1)"C"_2"H"_5"OH" + "HCl" → color(red)(4)"CrCl"_3 + "H"_2"O" + color(red)(2)"CO"_2 + "KCl}$

Step 4. Balance $\text{K}$

We have fixed 4 $\text{K}$ atoms on the left, so we need 4 $\text{K}$ atoms on the right. Put a 4 before $\text{KCl}$.

$\textcolor{red}{2} \text{K"_2"Cr"_2"O"_7 + color(red)(1)"C"_2"H"_5"OH" + "HCl" → color(red)(4)"CrCl"_3 + "H"_2"O" + color(red)(2)"CO"_2 + color(blue)(4)"KCl}$

Step 5. Balance $\text{Cl}$

We have fixed 16 $\text{Cl}$ atoms on the right, so we need 16 $\text{Cl}$ atoms on the left. Put a 16 before $\text{HCl}$.

$\textcolor{red}{2} \text{K"_2"Cr"_2"O"_7 + color(red)(1)"C"_2"H"_5"OH" + color(purple)(16)"HCl" → color(red)(4)"CrCl"_3 + "H"_2"O" + color(red)(2)"CO"_2 + color(blue)(4)"KCl}$

Step 6. Balance $\text{O}$.

We have fixed 15 $\text{O}$ atoms on the left and 4 $\text{O}$ atoms on the right.

We need 11 more $\text{O}$ atoms on the right. Put an 11 before $\text{H"_2"O}$.

$\textcolor{red}{2} \text{K"_2"Cr"_2"O"_7 + color(red)(1)"C"_2"H"_5"OH" + color(purple)(16)"HCl" → color(red)(4)"CrCl"_3 + color(brown)(11)"H"_2"O" + color(red)(2)"CO"_2 + color(blue)(4)"KCl}$

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

Step 6. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l)bb "On the right}}$
$\textcolor{w h i t e}{m m} \text{4 K"color(white)(mmmmm) "4 K}$
$\textcolor{w h i t e}{m m} \text{4 Cr"color(white)(mmmmll) "4 Cr}$
$\textcolor{w h i t e}{m l l} \text{15 O"color(white)(mmmml) "15 O}$
$\textcolor{w h i t e}{m m l} \text{2 C"color(white)(mmmmm) "2 C}$
$\textcolor{w h i t e}{m l l} \text{22 H"color(white)(mmmml) "22 H}$
$\textcolor{w h i t e}{m l l} \text{16 Cl"color(white)(mmmm) "16 Cl}$

The balanced equation is

$\textcolor{b l u e}{\text{2K"_2"Cr"_2"O"_7 + "C"_2"H"_5"OH" + "16HCl" → "4CrCl"_3 + "11H"_2"O" + "2CO"_2 + "4KCl}}$