The unbalanced equation is
"C"_2"H"_2 + "O"_2 → "CO"_2 + "H"_2"O"C2H2+O2→CO2+H2O
Let's start with the most complicated formula, "C"_2"H"_2C2H2. We put a 1 in front of it.
color(red)(1)"C"_2"H"_2 + "O"_2 → "CO"_2+ "H"_2"O"1C2H2+O2→CO2+H2O
We have fixed 2 "C"C atoms on the left, so we need 2 "C"C atoms on the right. We put a 2 in front of "CO"_2CO2.
color(red)(1)"C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + "H"_2"O"1C2H2+O2→2CO2+H2O
We have also fixed 2 "H"H atoms in the "C"_2"H"_2C2H2, so we need 2 "H"H atoms on the right. We put a 1 in front of the "H"_2"O"H2O.
color(red)(1)"C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + color(blue)(1)"H"_2"O"1C2H2+O2→2CO2+1H2O
Now we have fixed 5 "O"O atoms on the right, so we need 5 "O"O atoms on the left.
Oops! We would need to use 2½ molecules of "O"_2O2.
To avoid fractions, we multiply all coefficients by 2 and get
color(red)(2)"C"_2"H"_2 + "O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O"2C2H2+O2→4CO2+2H2O
We now have 10 "O"O atoms on the right, so we need 10 "O"O atoms on the left. We can now put a 5 in front of the "O"_2O2.
color(red)(2)"C"_2"H"_2 + color(orange)(5)"O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O"2C2H2+5O2→4CO2+2H2O
The equation should now be balanced. Let's check.
"Atom"color(white)(m) "On the left"color(white)(m) "On the right""AtommOn the leftmOn the right
stackrel(—————————————)(color(white)(m)"C"color(white)(mmmmm)4color(white)(mmmmmm) 4)
color(white)(m)"H"color(white)(mmmmm) 4 color(white)(mmmmmm)4
color(white)(m)"O"color(white)(mmmml) 10color(white)(mmmmmll) 10
The equation is balanced!
