 Okay, so let's try this one. It's a percent yield question. It says a sample of lime calcium oxide weighing 69.0 grams was prepared by heating 131 grams of limestone, which is calcium carbonate What was the percent yield of this reaction? So I have taken the liberty to write down the reaction So calcium carbonate solid is heated up To form calcium oxide and carbon dioxide goes away. Okay, so we're left with 69.0 grams of calcium oxide So once let's define the percent yield, but since calcium oxide and calcium carbonate Way different amounts. There are molar masses different. We're going to have to calculate the number of moles and then compare that okay, so Calcium It is 40.01 if I remember correctly Plus oxygen which is 16, so we've got 56 0.01 grams of calcium oxide There's one mole of calcium oxide so Okay, so I get 1.23 moles Calcium oxide, so that's how much we produced. This is how much was produced in the reaction How much could we have so what was the theoretical yield? Okay, so this is the produced or actual So what's the theoretical yield? Well, we've got to go from the starting material to calcium oxide. Okay, so for one mole calcium carbonate Well, let's calculate So what do we say? 40.01 plus 12.01 plus 3 times 16 100.02 Calcium carbon like that right so that would give us the mass of calcium carbonate But we want to know what the mass of calcium oxide would have been if it all reacted so that theoretical yield So where do we get that molar ratio? We look over at the reaction equation So for every one mole of calcium carbonate goes to one mole of calcium oxide so 31 divided by 1.02 And I get 1.31 moles of calcium oxide. So this is what we call So the percent yield Remember it's the part divided by the total so in this case it's going to be the actual divided by The theoretical And this would be the number of moles Thanks, and now we're comparing the same thing moles of calcium oxide to moles of calcium oxide So we got 1.23 moles of calcium oxide divided by 1.31 moles of calcium oxide Moles of calcium oxide canceled and so 1.23 divided by 1.31 times 100% If we go to three significant figures we get 93.9%, I think if we look on the paper that answers that. Any questions about that one? No. Hopefully it makes sense. It really does. I'll get it.