Question #cf542

1 Answer
Mar 15, 2016

Answer:

The balanced equation is #"2KOH"color(white)(l) + "CO"_2 → "K"_2"CO"_3 + "H"_2"O"#

Explanation:

The procedure I usually use to balance equations like this is:

  1. Start with the most complicated formula.
  2. Balance all atoms other than #"H"# and #"O"#.
  3. Balance #"O"#.
  4. Balance #"H"#.
  5. If necessary, multiply by an integer to remove fractions.
  6. Check that all atoms are balanced.

Your unbalanced equation is

#"KOH"color(white)(l) + "CO"_2 → "K"_2"CO"_3 + "H"_2"O"#

Step 1. Start with the most complicated formula.

Put a 1 in front of #"K"_2"CO"_3#. This number is fixed and does not change until (if necessary) Step 5.

#"KOH"color(white)(l) + "CO"_2 → color(red)(1)"K"_2"CO"_3 + "H"_2"O"#

Step 2. Balance #"K"#.

We have fixed #"2 K"# on the right, so we need #"2 K"# on the left.

Put a 2 before #"KOH"#.

#color(blue)(2)"KOH" + "CO"_2 → color(red)(1)"K"_2"CO"_3 + "H"_2"O"#

Step 3. Balance #"C"#.

We have fixed #"1 C"# on the right, so we need #"1 C"# on the left.

Put a 1 before #"CO"_2#.

#color(blue)(2)"KOH" + color(orange)(1)"CO"_2 → color(red)(1)"K"_2"CO"_3 + "H"_2"O"#

-+++++++++++++3
·······

Step 4. Balance #"O"#.

We have fixed #"4 O"# on the left and #"3 O"# on the right, so we need 1 more #"O"# on the right.

Put a 1 in front of #"H"_2"O"#.

#color(blue)(2)"KOH" + color(orange)(1)"CO"_2 → color(red)(1)"K"_2"CO"_3 + color(green)(1)"H"_2"O"#

All formulas have a coefficient, so the equation should be balanced.

Step 5.

(Not needed.)

Step 6. Check that all atoms are balanced.

#bb"On the left"color(white)(l) bb"On the right"#
#color(white)(mll)"2 K"color(white)(mmmmml) "2 K"#
#color(white)(mll)"4 O"color(white)(mmmmml) "4 O"#
#color(white)(mll)"2 H"color(white)(mmmmml) "2 H"#
#color(white)(mll)"1 C"color(white)(mmmmml) "1 C"#

The balanced equation is

#color(red)("2KOH"color(white)(l) + "CO"_2 → "K"_2"CO"_3 + "H"_2"O")#