you look at the equation on the elft side, the 02 on the left side is by iteslf, which would make it easiest to balance. It is also an even number. Now, if you look at the right side, the CO2 and H2O have 3 Oxygens. Since 3 is an odd number, you can start by multiplying H2O by 2. Now you have 4x Oxygen on the right side. By a new problem occured, you now have 4x Hydrogen on the right side in comparison to the 2 on the left. C2H2 + O2 -- CO2 + 2H2O How do you balance that? You multiply C2H2 by 2. Now the hydrogen are balanced but you have 4x Carbon on the left side and 1 on the right side. 2C

Question

you look at the equation on the elft side, the 02 on the left side is by iteslf, which would make it easiest to balance. It is also an even number. Now, if you look at the right side, the CO2 and H2O have 3 Oxygens. Since 3 is an odd number, you can start by multiplying H2O by 2. Now you have 4x Oxygen on the right side. By a new problem occured, you now have 4x Hydrogen on the right side in comparison to the 2 on the left. C2H2 + O2 --> CO2 + 2H2O How do you balance that? You multiply C2H2 by 2. Now the hydrogen are balanced but you have 4x Carbon on the left side and 1 on the right side. 2C