The procedure I usually use to balance equations like this is:
- Start with the most complicated formula.
- Balance all atoms other than "H" and "O".
- Balance "O".
- Balance "H".
- If necessary, multiply by an integer to remove fractions.
- 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")
