Let's run through a stoichiometry example of doing a combustion reaction. At this point, if you want to see the question, pause the video, and then solve it on your own, then you can use the rest of the video to check your work. Let's talk about the combustion of 50 grams of propane gas, and we're going to burn that with plenty of oxygen so that the reaction goes to completion. I would like to know what mass of water is produced, what mass of carbon dioxide is produced, and how much oxygen was consumed in the process. Propane is C3H8, and in the combustion reaction, we're going to react it with oxygen, and assuming that we have sufficient oxygen, I'm going to make carbon dioxide and water. I have to balance this, so I'm going to put three carbons over here. I need eight hydrogen, so I'm going to put a four over here. I have six oxygens in the carbon dioxide and four oxygens in the water, so that's 10 oxygens, so I have to put a five in front of the O2. So there's my balance equation, step one. The next step is I have to convert to moles. I'm going to take my 50 grams of C3H8, and then I'm going to look up the molar mass of propane on the periodic table. When I add up the mass of three carbons and eight hydrogens, I find that one mole is equivalent to 44.0962 grams. Cancel out the grams, and I'm left with 1.13 moles of propane. All right, so that was step two, convert to moles. Now that I'm in moles, I can use the ratios for my balanced equations to predict how much product I make. So I have 1.13 moles of C3H8. If I want to find how much carbon dioxide I make, there are three carbon dioxide for every one C3H8. Cancel out the C3H8, and I get 3.40 moles of CO2. That's step three. Well, the question asked for the mass of carbon dioxide. So if I have 3.40 moles of carbon dioxide, the periodic table tells me there are 44.00 9 grams for every one mole. And so I produce 150 grams of CO2. So step four is to take our data and convert it back to grams. Written about the equations, convert our data to moles, use the ratio from the balanced equations, and convert our data back to grams. Now we can run through the same thing for water. We've already balanced equation and convert to moles. So for water, I'm going to say I have 1.13 moles of propane. Now instead of being a 3 to 1 ratio, there are four waters for every one propane. So 1.13 times four, I have 4.52 moles of water. Now I can convert that to grams. I can say 4.52 moles of water. And from the periodic table, I know that water has a molar mass of 18.0148 grams for every one mole. So I get 81.4 grams of water. To find the mass of oxygen, I simply need to do the conservation of mass. I know that I started with 50 grams of propane. I know I make 150 grams of carbon dioxide and 81.4 grams of water. So that means I produce 231.4 grams of stuff, which means I need to start with 231.4 grams. So if I start with 231.4 grams and only 50 grams of it is propane, that means I must have 181.4 grams of oxygen.