 So let's try a real example. We're going to perform this reaction here, lead nitrate and potassium iodide, reacting to give potassium nitrate and a bright yellow precipitate of lead iodide. And we're told that 0.831 grams of that precipitate, the lead iodide, is produced and we are to work out what mass of potassium iodide must have reacted. So we have the balanced equation already. Now let's figure out what calculations are needed. Well we're dealing with lead iodide and potassium iodide and we have 0.831 grams of lead iodide. So the steps of the calculation are we need the molar mass of lead iodide. We'll then use that to work out the moles of lead iodide. We'll use the mole ratio from the equation to work out how many moles of potassium iodide must have reacted. We'll then need the molar mass of potassium iodide and then we'll use that to convert the moles of potassium iodide to a mass. You might like to pause the video now and see if you can work through those calculations yourself. So let's work through those steps. First we'll calculate the molar mass of lead iodide. Look the information up on the periodic table. Lead is 207.2 grams per mole and iodine is 126.9 grams per mole. So the molar mass you should get is 460.8 grams per mole for PBI2. Now let's use this to calculate the moles of lead iodide. So we take 0.831 grams of lead iodide, put it over one and divide by the molar mass. Remember that dividing is the same as multiplying by the inverse. So that puts the grams on the bottom and allows us to cancel those units out. And this calculation gives us 0.0018304 moles of lead iodide. Notice that at each step I'm writing down exactly what I just calculated. This is what I mean by annotating your calculations. This will prevent me from mixing up values later on and makes it easier to spot or track down mistakes when you're checking your work. Okay, now we know the moles of lead iodide. We can use the mole ratio from the equation to work out how many moles of potassium iodide must have reacted. You can see that the ratio is one lead iodide to two potassium iodides. So we need to multiply the moles of lead iodide by two. And that gives us 0.0036068 moles of potassium iodide. Notice also that I'm maintaining plenty of sig figs as I work through the calculation. If you round off during a calculation you can introduce rounding errors that distort your final value. If you don't want to write down all the digits on the calculator just make sure you write down more sig figs than you'll need in your final answer. You can see because we started with 0.831 grams our final answer is going to be rounded to three sig figs. So I'm keeping at least five in my calculations until the final answer. Okay, last step. We have the moles of potassium iodide. We just need to convert that to a mass. For that we need the molar mass of potassium iodide. So again we consult the periodic table and we work out that the molar mass of Ki is 166.0 grams per mole. Now we convert the 0.003608 moles of Ki to a mass by multiplying by the molar mass. This time the moles cancel out leaving grams. Which gives us 0.59873 grams which rounds to 0.599 grams of potassium iodide. And that's our final answer.