How do you balance an equation?

The best way to go about this, and all balancing, is to first balance the element that appears only ONCE on each side. In this case, that would be carbon. So, since there are 6 carbons on the right, there has to be 6 on the left. So you get H2O + 6CO2 ---> C6H12O6 + O2

Next, balance the hydrogens. On the right you have 12 hydrogens; on the left you have 2 hydrogens. Make 12 hydrogens on the left by adding 6 H20. Now you have 6H2O + 6CO2 ---> C6H12O6 + O2. But now you have 18 oxygens on the left and 8 oxygens on the right. To make 18 oxygens on the right you'd can add 6O2. That now gives you 6H20 + 6CO2 ---> C6H12O6 + 6O2.

In summary, you now have 6 carbons on the left and 6 carbons on the right; you have 12 hydrogens on the left and 12 hydrogens on the right; and you have 18 oxygens on the left and 18 oxygens on the right.

I know this seems rather circuitous, but that's the way one goes about it. The best advice is to found an element that only appears once (in one reactant and in one product) and balance that first. If oxygens and hydrogens (as in water) appear, do the hydrogens next, and then finish up with they oxygens.

H2O + CO2 ---> C6 H12 O6 + O2

So to balance it we need the same number of each element on both sides (since nothing is lost in a balanced equation)

I always start with the least common element in the equation, so here it's Carbon. There are 6 carbons on the right, so there need to be 6 on the left:

H2O + 6CO2 ---> C6 H12 O6 + O2

Same goes for the Hydrogens:

6H2O + 6CO2 ---> C6 H12 O6 + O2 <(remember Hs come in pairs so you put 6 not 12)

But since the Cs on the left are included in carbon dioxide, there are now 6 O2s (12 oxygens in total). The same with hydrogen, there are now 6 Os.

Therefore in your unbalanced equation, we have worked out that on the left side there are 18 oxygen molecules, but only 8 on the right.

There are already 6 included in the carbohydrate, so there need to be 12 more Os (or 6 O2s) on the right hand side:

6H2O + 6CO2 ---> C6 H12 O6 + 6O2

Now count the elements on both sides = should be balanced perfectly :)

6H2O + 6CO2 --> C6H12O6 + 6O2

Count the molecules of both sides. On the left side, there are 2 Hydrogen, 1 Oxygen, 1 Carbon, and 2 Oxygen, which is 2 Hydrogen, 1 Carbon, and 3 Oxygen. The other side has 6 Carbon, 12 Hydrogen, 6 Oxygen, and 2 Oxygen, which adds up to 6 Carbon, 12 Hydrogen, 8 Oxygen. Put a 6 in front of the H2 on the left side to get 12 Hydrogen on both sides. This will give you 6 Oxygen also. Put a 6 in front of CO2 to get 6 Carbon and 12 Oxygen. Everything else is balanced except for the Oxygens. There are now 18 on the left side and only 8 on the right. Put a 6 in front of O2 on the right side to get 12 O + 6 O= 18 Oxygen. Both sides are now balanced.

Chill out.