Question

Can anyone explain to me how to work stoichiometry problems?

Answer 1

You write an equation and you balance it.

Answer 2

\[C_{6}H_{12}O_{6}+O_{2}\rightarrow H_{2}O+CO_{2}\]
Ok let's balance this sample problem. The main goal is to get the same amounts of carbon, hydrogen, and oxygen on both sides. Let's just tally them up here and see what it looks like:

Left/Right
6 Carbon/1 Carbon
12 Hydrogen/2 Hydrogen
8 Oxygen/3 Oxygen

So it looks like we have a lot to balance, but as long as you set your priorities straight you'll be fine.

You should find what molecules need to be balanced that are by themselves and balance those last. In this case it's oxygen because on the left side it appears as O2 all alone. That way when you're done you just give it whatever it needs to balance out. If you save H2O or CO2 for last, then you'll end up messing something up in hydrogen or carbon. If this seems weird, just look at what I do.

Balance carbon and then hydrogen
\[C_{6}H_{12}O_{6}+O_{2}\rightarrow 6H_{2}O+6CO_{2}\]

Let's look at what we have again

Left/Right
6 Carbon/6 Carbon
12 Hydrogen/12 Hydrogen
8 Oxygen/18 Oxygen

Now we see that to make oxygen on the left equal to the right that we have

6+2x=18

This shows that the left and right must be equal. The 6 comes from C6H12O6 and the 2x comes from our unknown variable in front of O2. Since if we put a 3 in front of O2 we're really adding 6 oxygen, that's where 2 comes in. And of course, the left side all equals 18.

Solve it and you get x=6 for the coefficient

\[C_{6}H_{12}O_{6}+6O_{2}\rightarrow 6H_{2}O+6CO_{2}\]

Count them up one last time to check:
Left/Right
6 Carbon/6 Carbon
12 Hydrogen/12 Hydrogen
18 Oxygen/18 Oxygen

And that's a balanced equation.