Let's look at a problem that combines some of the important topics in general chemistry. When lead to sulfide or burns in oxygen, the products are solid lead to oxide and sulfur dioxide gas. Let's break this question down into two parts. The first part is asking you to write and predict the products of a chemical reaction. The second part of the problem is the tricky part. Based on the chemical equation you just wrote, can you determine how much oxygen is needed if you have 30.0 grams of lead to sulfide? Think of the problem like this. Let's say you're trying to bake a cake and the recipe calls for one cup of sugar, half a cup of butter, two eggs, one and a half cups of flour, and half a cup of milk. Remember that that's the amount of stuff you need in order to make one cake. How much cake do you make if you cut all of your ingredients in half? Well, that's right. You only make half a recipe or half a cake. What happens if you multiply all the ingredients by three? You would be able to make three cakes. Now here's the challenging part. What if you start with the original recipe, but instead of having two eggs, you just have one egg. How many cakes can you make? Because the amount of eggs is cut in half, that limits how much cake you can make at one time. In chemistry, we call an ingredient that limits how much product can be made a limiting reactant. In our problem, we have 30.0 grams of lead to sulfide, so how much product do we make? And how much oxygen do we need to make it? This is the heart of the problem we're dealing with. Let's look at the first part, write the balance chemical equation for the reaction. We know what our reactants are. PBS solid plus O2 gas. The products are solid lead to oxide and sulfur dioxide gas. Putting both sides of the equation together, we have PBS as in parentheses plus O2 G in parentheses yields PBO as in parentheses plus S02 G in parentheses. We often show the physical state of a substance using parentheses as for solid G for gas and alpha liquid. Before we're done, we have to make sure that this equation is balanced is this equation balanced. Well, we have one PB on each side and we have one as on each side, which are balanced. But we have two oxygen in the left side and three on the right side, which are not balanced. To balance this out, the coefficient in front of the O2 should be three over two because the three comes from the right side and two comes from the left side. But we don't like having fractions in the left side. So we have to multiply three house by two so that we just have three as the coefficient. Since we want to multiply that coefficient by two, we have to multiply both sides of the equation by two. So now the complete balance equation becomes two PBS as in parentheses plus three O2 G in parentheses yields two PBO as in parentheses plus two S02 G in parentheses. This is our final answer for part A. Now let's talk about part B, which asks how many grams of oxygen are required to react with 30.0 grams of lead to sulfide. What do we know about this problem? We found the balanced equation in part A and we know we're starting with 30.0 grams of lead to sulfide. What do we want to find out? We want to know how many grams of oxygen we need to react with 30.0 grams of lead to sulfide. The next step is how do we go from grams of lead to sulfide to grams of oxygen gas. The plan of attack is to use the molar mass of lead to sulfide to convert grams of lead to sulfide to moles of lead to sulfide. Once we're at moles of lead to sulfide, we can convert to moles of oxygen gas because we know the chemical equation and the relationship between lead to sulfide and oxygen gas. Once we're at moles of oxygen gas, we can convert back to grams of oxygen gas by using the molar mass of O2. So we need to find three conversion factors. One, how many grams of lead to sulfide are a one mole of lead to sulfide? When we calculate the molar mass, we find that one mole is equal to 239.3 grams of lead to sulfide. Two, how do we convert from moles of PBS to moles of O2? The equation that we saw for in part A told us that for every two moles of PBS we have, we need three moles of O2 in order to make two moles of lead to oxide. So our conversion factor is two moles of PBS is equal to three moles of O2. Three, how many grams of O2 are a one mole of O2? This is just like the first conversion factor. One mole of oxygen gas is equal to 32.0 grams of oxygen gas. With these conversion factors, let's use dimensional analysis to find the answer to part B. Starting with 30.0 grams of lead to sulfide, we use the molar mass of lead to sulfide to write the first conversion factor as dividing one mole of lead to sulfide over 239.3 grams of lead to sulfide so the grams will cancel out. Now that we're in moles of lead to sulfide, we can multiply by the next conversion factor, which is every three moles of oxygen gas equals two moles of lead to sulfide to cancel out moles of lead to sulfide. Since we need to find grams of oxygen gas, the next conversion factor is 32.0 grams of oxygen gas over one mole of oxygen gas. So moles of oxygen cancel out and we're left with our final unit, grams of oxygen gas. Mathematically, this is 30.0 times 1 times 3 times 32.0, all divided by the quantity 239.3 times 2 times 1, which gives us 6.02 grams of oxygen gas as the final answer.