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jb

Jilan B.

Calculus 3

8 months, 3 weeks ago

When hydrobromic acid (HBr) reacts with chlorine gas, bromine and hydrochloric acid are obtained: 2($% (,-) + /0#(1) ? $%#(0) + 2(/0(,-) i) When 17.1 g of hydrobromic acid reacts with 46.2 g of chlorine, what is the theoretical yield of hydrochloric acid? ii) If the actual yield of hydrochloric acid is 4.23 g, what is the percentage yield?

Indiana University Bloomington

Answer

When hydrobromic acid (HBr) reacts with chlorine gas, bromine and hydrochloric acid are obtained: 2($% (,-) + /0#(1) ? $%#(0) + 2(/0(,-) i) When 17.1 g of hydrobromic acid reacts with 46.2 g of chlorine, what is the theoretical yield of hydrochloric acid? ii) If the actual yield of hydrochloric acid is 4.23 g, what is the percentage yield?

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Video Transcript

Yeah. Hi there in this problem, we have two questions actually. We are trying to calculate theoretical yield and then use that theoretical yield to calculate percent yield. Um So calculating a theoretical yield is a strike geometry problem. We're going to have to determine the number of grams of product we can make given the grams of each reactant. So as in any strike geometry problem, the first thing we need is a balanced equation. So we have hBR reacting with chlorine and when chlorine is by itself, not combined with anything else, it always exists as the molecule Cl two. We're going to have a single replacement reaction here. It is going to form hcl and brahmi grooming is another one of those elements that when it's by itself, it's the diatonic molecule BR two. So to balance this, I see that I have to chlorine is on the reactant side and only one on the product side. So I'm going to have to put a chew there that gives me two hydrogen. So, coming back over to my reactant side, I'll put a two in front of that and then that gives me to be ours and I have to be ours. So this is now balanced. In part I we are given 17.1 g of H p r and 46 0.2 g of chlorine. And we are trying to determine the theoretical yield the number of grams of hcl that will be produced. All right. To to determine the number of grams of Hcl that can be produced. We are going to have to do strike geometry. But before we can do that, we need to figure out which of these two reactant is are limiting reactant. Remember, a limiting reactant will determine how much product you make. It is the one you are going to run out of first. So we need to figure out if the 17.1 g of HBR is limiting. Or if the 46.2 g of cl two is limiting because the limiting reactant is the one I want to use to calculate the grams of hcl. So to do that, I am going to need to convert both of these two moles because I don't know the graham ratio these react in. But I do know the mole ratio because my balanced equation, I know that H BR and Cl to react in a 2 to 1 ratio for every two most of HPR, I need one mole of cl two. All right. So, what I want to do first is figure out which of these is limiting. So, I need to convert both of these two moles. So, I have 17.1 g of H p R. And to convert that to moles, I need to divide by smaller mass. So looking at the periodic table, I take the mass of hydrogen and the mass of VR and at those together. And I get 80.91 grams of H p R in every mole of HBR Yeah, grams of HBR will cancel. So if I take 17.1 and divided by 80.91 I get 0.211 moles of HBR. All right, let's go through the chlorine and figure out how many moles of chlorine we have. Starting with 46.2 g of chlorine. Okay, one mole of chlorine. I need to take the molar mass of chlorine and multiply by two since it's cl two. So that gives me 70.90 g of cl two. Yeah, he had 0.652 moles. Okay. All right, so this is how many moles of these we have We know from the balanced equation, but they react in a 2 to 1 ratio. That means I need twice as much HBR in terms of moles as cl two. Well, looking at these, I don't even have an equal amount of the two. I have far less HBR So I do not have twice as much HBR as I have Cl two. So, that means the HBR is my limiting reactant. That is the one I am going to run out of first. That is the one that is going to determine how much hcl we can make theoretically. So that is the amount that I'm going to use. Moving forward to figure out my theoretical yield of Hcl. All right, so, I'm going to do a little bit of striking geometry here. Starting with that point to 11 moles of my limiting reactant. HBR first thing I'm going to use is the mole ratio from the equation showing me how many moles of h c l. I can make per mole of HBR, there's a coefficient of two in front of the HBR so for every two moles of H p r, I can make two moles of Hcl because there's also a two in front of the hcl in the balanced equation. So you might be asking why did we have to do this step? If it's just to over two, it's just it's not going to change the number. The answer to that is it converted our units. So this step is very important to get us two moles of Hcl because now we can take the molar mass of hcl and multiply by that T eight g of hcl that are produced and the solar masses 36 0.46 g of hcl in every mole of hcl moles of hcl cancelled. And I have the unit grams of hcl in my answer. Specifically 7.71 g. This is our theoretical this is how much we should make when we start with 17.1 g of HBR So theoretical is the amount predicted by math. Okay. The amount that should be made. All right. So this is our first answer. This is the answer for part I 7.71 g of hcl. Moving on to the second part of the question. The second part of the question gives us an actual yield and wants us to calculate percent yield. So we need the formula for percent yield, percent yield is defined as the actual yield. But you actually get when you do the experiment in the lab divided by that theoretical yield. The amount you calculated by doing so like geometry, that's the expected or what you think you should get. But the actual is what you actually get. Then we are going to change this to a percent by taking that decimal that we get and multiplying by 100 and adding the percent sign. So the actual yield given to us in the problem is 4.23 g theoretical yield is 7.71 g, dividing those numbers, changing the decimal 2% by multiplying by 100 I get 54 0.9 percent as the percent yield. That is the answer for the second part of the question. Okay thank you for watching.