Both Jacques Charles and Joseph Louis Guy-Lussac were avid
balloo...

Question

Both Jacques Charles and Joseph Louis Guy-Lussac were avid
balloonists. In his original flight in 1783, Jacques Charles
used a balloon that contained approximately 31,150 L of H2.
He generated the H2 using the reaction between iron and hydrochloric
acid:
Fe1s2 + 2 HCl1aq2 ¡ FeCl21aq2 + H21g2
How many kilograms of iron were needed to produce this
volume of H2 if the temperature was 22 °C?

Accepted Answer

Hey there, folks, welcome back. Alright, So one of the common commercial bleaching agents is Glory in dioxide C. L. 02. It bleaches materials by oxidizing them while itself it gets reduced. Um chlorine dioxide can be prepared by the reaction of chlorine and sodium chloride. As we have in this reaction below, we want the mass of cielo two of chlorine dioxide. So, we want mass of this. They can be prepared when we have 18.6 g of sodium chloride. Um that's allowed to react with three liters of chlorine gas. All right. So, we actually have um starting information for both of the reactant. And you know what that means When we're giving amounts of both of the reactant, we need to find which one is the limiting reactant because that's the one that's going to actually give us the accurate theoretical amount of the product. Right? So even though for the one of them we have that ingram's the other one. We don't have that ingram's we're told that we have three liters of chlorine gas, which means we're gonna have to use the ideal gas law in order to um, you know, basically find the number of moles of cl two that we have. But let's go ahead and start with the easy one with 18.6 g of sodium chloride. We are going to have to calculate it smaller mass first. Right? So, we have 18.6 g sodium chloride. So, in one mole of this compound, how many grams are there? That's what we're going to go ahead and figure out before we can convert that into moles. All right, so, we have sodium sodium. Let's see if you take a look at the product table, Where is the sodium weighs 23g? We have chlorine. Just one and it weighs 35.45 g. And then oxygen. We have two of them. Here they weigh 16 g each. Oops. And that will be 32 g. So go ahead and add those three numbers together. And that will be 90.45. Which is the molar mass Of this compound. Okay, so in one mole of sodium chloride there are .45g of sodium chloride. So we've canceled out the grams of sodium chloride. Now we have moles and now we can actually figure out the moles of the chlorine dioxide. Right? Because we have a chemical equation, we just need to make sure that this is a balanced equation. Let's go ahead and check really quick to chlorine is to sodium Um to oxygen. Yeah, we're good. So this is balanced. Always make sure you check just in case. All right. So now we can actually do a multiple comparison between um sodium chloride and chlorine dioxide. Right? So, we can say that it's actually a 2-2 mole ratio. So every two moles of NAC. L. two used to produce two moles of C L. 02. So, we put moles of N A C. L. 02 on the bottom because we want them to cancel out. So now we have moles of that. Let's go ahead and figure out how many moles that is Right. Once you divide and multiply everything here, we're going to get . moles of C L 02. Obviously that's not going to be our final answer. This is this is a possible amount that we can produce. Now. We need to take a look at the chlorine gas. So we have three liters of that. So this is going to be a little bit different. We're going to go ahead and take a look at um the ideal gas law equation right PV equals NRT. So what we're going to try to do here is actually solve for N of chlorine gas because for the pressure we're giving the pressure, we're given, the temperature were given the volume. Um our hair is always given because it's a constant. So the only thing that we're missing here is N. Right, so let's go ahead and solve for end the bible sized by R. T. So we have P. V over R. T. And now we can plug in the numbers of course we do need to make sure all of the units are correct. So for pressure should be in atmospheres and we have atmosphere. So 2.2 That's correct volume should be in leaders and it isn't leaders three leaders Alright are here is just a constant. You should remember the value along with the units . 206 leaders times atmospheres over mold times Calvin. Okay, so that's our and then temperature here has to be in Calvin's as well. If you don't remember, just take a look at the units of the gas constant. We have Calvin in there. Do we have Calvin? No, we have 23°C. So just go ahead and add to 73.15 to the C. And you're going to get Calvin in return. So that will give us To 96.15 Calvin. All right, So to 96. Calvin's. Alright, so let's go ahead and multiply the top numbers, multiply the bottom numbers and go ahead and um divide them together. Notice that atmospheres cancel out. So do the leaders and the Calvin. And we're going to have moles left over here as the unit because we're solving for moles. Okay, so moles here moles of of cl two. Right, so this is for C. O to not cielo too. Alright, so the number there will be given just a second . moles of C 02. Now that we have moles. So we had to do all of that just to find moles so that we can do a multiple comparison between the two and find how many moles of cielo to that cl two can produce. Alright, so we're going to go on the bottom here And we're going to say that .2716 molds of steel too. We're going to go ahead and do a multiple comparison. What is the mall to mall comparison? So it is one mole of Cl two for every two moles of Cielo too. So one more of CO2 for every two moles of C L 02. And that cancels out the moles of cl two gas and gives us the product amount. And that number should be . moles. Alright, so we need to take this number and we need to take this number that we found and compare them. So which one is actually going to be the correct mole amount? So, the one that we found the first is actually a smaller number. Right? Why is that? Well, because it's being produced by a limiting reactant. So NACL 02 is limiting. Yeah. So I changed the pen color. So this one is going to be our limiting reactant and that's why it's producing the least amount of the product, even though CL two is able to and wants to produce more, it can't because once the limiting reactant is used up, the reaction stops. Right? So this is actually not the correct amount of the product that we're going to use um even though skill to wants to do that. So this is the amount that we're going to take a look at now, what we want here is the mass of Cielo to not moles. Right? So we just need to do one more step here. Let me take that out. So this is the amount that we're going to use and we're gonna go ahead and just convert that into grams. So in one mole of C L 02, How many grams are there? Alright, so we have, I'm gonna put it right here. So C L 02. So we have chlorine. Just one. Right? And it's 35.45 g per mole. Then we have oxygen to oxygen's so times 32 g per mole And then the total maps will be 67.45. Okay, so in one mole there are 67.45. Oops, not 49 45 grams of Cielo too. Alright, so moles cancel out and that will give us a number and grams the answer in grams. And let's see that will be 13.86 g of C L. 02. I'm gonna go ahead and highlight this number now because that is the answer that they wanted. All right. So here we were given a balance the reaction. We were given amounts of both of the reactant, although in different forms, one was in mass, one was in volume. So, we actually had to do some extra steps to find the moles of C. 02 from the, from the volume from the temperature and pressure by using the ideal gas law. Right? So that's why it took us a little bit longer. But once we had the molds of Cl two, we just did regular stock geometry, multiple comparison, and then we had the moles of of the products that were being produced by both of the reactant to remember the one that produces the least amount is limiting and that's the number that we want to go with. So that's why we used the molds of CL two from, from the sodium chloride reactant and not from seal to reactant. Alright, alright, folks, thank you so much for watching and we'll see you in the next video.

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