Question #64b84

1 Answer
Apr 13, 2017

Answer:

Warning! Long answer.

#"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.

We start with the unbalanced equation:

#"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:

#"Cr: +6 → +3; Change ="color(white)(lll) "-3 (reduction)"#
#"C:"color(white)(ml) "-2 → +4; Change = +6 (oxidation)"#

Step 2. Equalize the changes in oxidation number

We need 2 atoms of #"Cr"# for every 1 atom of #"C"# or 4 atoms of #"Cr"# for every 2 atoms
of #"C"#.

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

Step 3. Insert coefficients to get these numbers

#color(red)(2)"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 #"K"#

We have fixed 4 #"K"# atoms on the left, so we need 4 #"K"# atoms on the right. Put a 4 before #"KCl"#.

#color(red)(2)"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 #"Cl"#

We have fixed 16 #"Cl"# atoms on the right, so we need 16 #"Cl"# atoms on the left. Put a 16 before #"HCl"#.

#color(red)(2)"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 #"O"#.

We have fixed 15 #"O"# atoms on the left and 4 #"O"# atoms on the right.

We need 11 more #"O"# atoms on the right. Put an 11 before #"H"_2"O"#.

#color(red)(2)"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.

#bb"On the left"color(white)(l)bb "On the right"#
#color(white)(mm)"4 K"color(white)(mmmmm) "4 K"#
#color(white)(mm)"4 Cr"color(white)(mmmmll) "4 Cr"#
#color(white)(mll)"15 O"color(white)(mmmml) "15 O"#
#color(white)(mml)"2 C"color(white)(mmmmm) "2 C"#
#color(white)(mll)"22 H"color(white)(mmmml) "22 H"#
#color(white)(mll)"16 Cl"color(white)(mmmm) "16 Cl"#

The balanced equation is

#color(blue)("2K"_2"Cr"_2"O"_7 + "C"_2"H"_5"OH" + "16HCl" → "4CrCl"_3 + "11H"_2"O" + "2CO"_2 + "4KCl")#