In this video, we're going to be solving the story geometry problems. The first step to any store archaeology problem, did you write the balanced chemical reaction? So this would be our balanced chemical reaction between before the reaction between chlorine gas and sodium metal. We want to know how many moles and how many grams of the chlorine Gas is needed to react with 10 g of sodium metal. So what we were going to do is, first of all, we need to make sure our equation is balanced and now it is. So what we have to do is start with the value that were given. So that's the massive sodium. And then since we have the massive sodium and we want to convert to the moles of sodium, what we will need to do is use the molar mass or atomic mass of sodium in this case, since it's an atom to convert two moles of sodium and then we're going to take the mole ratio from the balanced chemical reaction, which is two moles of sodium girls, one more of Corinne, And to get the mould of chlorine, we can stop right here. We just have to calculate this. It would just be 10 divided by 2.99 Times two. However, to convert to the mass of the chlorine, we need to use the molar mass of the chlorine gas taken out. There are two chlorine atoms. So we're going to multiply The atomic mass of chlorine by two and this is what we get. And then we would just solve that in order to get the mass of the chlorine. So it's just one more step for the massive chlorine. Okay, now let's do another problem. The next problem is asking us to find the bowls and massive oxygen that's formed by decomposition of mercury to oxide. So first of all, let's start with writing gap mercury too oxide. So the roman numeral indicates the charge on the mercury and oxygen always has a charge of negative two. So therefore this is mercury oxide. We're told that this is decomposing. So the composition of this is going to give us the following and we're going to balance this using coefficients and that's what we end up with. And then we are going to be Decomposing 1.2 5 2 g of mercury to oxide. Yeah. And we're looking for moles and mass uh, oxygen. So we're going to do is we have to follow the same process as in the last problem, we need to find the molar mass of mercury oxide. So we add up the atomic mass of mercury. Now the atomic mass of oxygen. And then this is what we end up with. Okay. And then using the story geometric coefficients from the balanced chemical reaction, we convert From moles of the Mercury Oxide, two moles of the oxygen. And this is how we get the moles. If we stop right there, however, we need the mass. We need to use The molar mass of oxygen, which is going to be 16 times too. So that's going to be 32 grams of oxygen per mole. And that's how we would get the mass of the oxygen. There was just one more step. Mhm. That's it. Mhm. Now, the next problem that we're going to do is we're going to find the number of moles and massive sodium nitrate required to produce 128 g of oxygen. So, the reaction that we're looking at right now is The sodium nitrate, which is written at NANO three sodium nitrate. It's going to give us two products so we have to balance this equation. Mhm. All right. So double checking the elements on both sides. We can see that this is now balanced. Now, we're told That 128 g of oxygen are produced. We want to know the moles and the mass of the sodium nitrate. That would be used to produce this amount of the oxygen. So therefore, let's start with what we're given. 128 g of oxygen. We use the more mass of the oxygen and then from the balanced chemical reaction, get the mole ratio. And then, as we did with the other problems afterwards, if we want to get the mass, then we have to use the molar mass, uh the sodium nitrate. And we can calculate that using the periodic table. And I'll let you do that on your own. No. The next problem we're going to do is find the number of moles and massive carbon dioxide formed by combustion of carbon with and access of oxygen. We're told the oxygen is an access so we know that carbon is going to be the limiting reacted this is his combustion. The carbon will react with oxygen to form carbon dioxide. So this is already balanced. We're told that we have 20 kg of carbon in order to use lower mass to convert the carbon too, from grams to moles. We have to convert kilograms two g in order to convert kilograms two g, multiplied by 1000. We move the decimal .3 places to the right. So that's going to be that make grams of the carving. Now, what we need to find is the moles and mass of the carbon dioxide. So this problem after converting, we will follow the same process. So we start with the mass of the carbon from periodic table. We obtain the atomic mass of carbon, Convert the carbon two moles from the balanced chemical reaction, Convert moles of carbon two moles of carbon dioxide. And if we want to get the mass of the carbon dioxide, then use the molar mass of carbon dioxide, which we obtained by adding up the atomic mass of carbon and to oxygen. And that's what we get. And solving that, we will get mass. Oh, carbon dioxide. Okay, now let's do another problem. We're going to find the moles and mask of copper to carbonate. Need to produce carbon to oxide. First of all, we need to write down the formula for copper to carbonate. Carbonate has a charge of negative too. So it's going to combine with one atom of the copper in formula. So that would be our formula. And then it's going to give us the copper to oxide. And that would be the following. The other products that we are going to get is carved been dioxide. So this is going to be our reaction. Now we need to check that this is balanced and we see that it is so we can move on to the next step. Now we're going to be producing 1.500 kg of the copper to oxide. So it was right that down here. Now we need to convert this two g. So again, multiply Bye 1000. And that's how many grams of the copper to oxide that we have. Now we need to find moles and mass of the reacted. So let's once again startup dimensional analysis like we have been doing again use the molar mass To convert two moles of the copper to oxide. And then use the mole ratio between the product and the reactant that we're looking for to convert two moles of the reactant. That's what we get. And in order to get from the moles of the reactant to mass of the reactant, use the molar mass, as we did with previous problems, and that would be how we would do that last problem.