A sample of 8.69 g of Zn1OH22 is added to 155.0 mL of 0.750 M H2SO4. (c) How many moles of ZnSO4 are present after the reaction is complete?

Video transcript

Hey there folks, welcome back. Alright so in this practice problem we have 200 ml of 0.300 moller silver nitrate and it's added to mL of 400.2 point 00 Mueller sodium bromide, we need to determine the mass of silver bromide solid that is produced. So first of all, we need to go ahead and write a balanced equation for this because we are going to be doing stoic geometry and the geometry of course balanced equations or everything. Alright, so we have sodium, I'm sorry, we have silver nitrate and sodium bromide reacting. So those are going to be our reactant. So let's go ahead and write them. So E. G. And oh three silver nitrate nitrates are all soluble. So of course it's going to be a quiz. Right then we have an A. B. R. sodium bromide and because we have a group on a um iron hair metal, it is also going to be Aquarius because it's soluble. Now we're going to produce here, we're told that we're producing silver bromide solid. Right? So let's go ahead and write that out silver bromide solid. And this is going to be our insoluble products because here we have a halogen roaming with either silver mercury or lead ion. Right, so that's why it's insoluble. But of course we are missing one product. Right so let's take a look at what we have. So if we go ahead and break down both of the reactant into their ions. So we have AD plus one and oh three minus and then we have N. A plus one, N. B. R minus. So as we can see on the product side we have silver plus one combined with B R minus one. And now the only thing that's left is sodium plus one and then oh three minus one nitrate. Right? So they're going to combine into any N. O. N. 03. And here again, nitrates are soluble so it's going to be Aquarius. So we have sodium nitrate here as the other products. All right so let's go ahead and check make sure everything is balanced. So we have the same number of sodium of silver browning. Um We can go ahead and look at N. 03 as as a poly atomic iron. It doesn't really change. So there's still just one mole of N. 03 minus here. So yes everything is balanced so that we can see everything is just 1212 mol ratio. Right? So let's go ahead and start with the 200 mL of silver nitrate. We're going to go ahead and use this information and figure out how much of um this product of a G. B. R. Three or I'm sorry, KGB are solid. What is the mass of this? That similar nitrate is going to produce? And then we're going to figure out how what's the mass mass that sodium bromide is going to produce of that product and then whichever is going to produce the least amount. That's obviously going to be our answer because theoretical um answers Or theoretical yield is going to be produced by the limiting reactant recall that. Alright, so let's go ahead and start with the 200 milliliters. Now, if we take a look at polarity here, remember polarity big M is moles over a leader. Right? And here we have milliliters. So first before we are able to multiply the volume with the modularity and we actually need to go ahead and convert this militia leader into a leader. Okay, so miller leader will go on the bottom and leader on top million metric prefixes gets one in front of that and millie is going to be 10 to the negative three. Right? So there is basically uh 10 to the negative three liters and one millimeter. Okay, So now we can cancel out the ml and now we have leaders. Now we can go ahead and multiply by the polarity which is .300. And instead of polarity I'm just going to write moles per one Leader because it's the same thing. And this is for AGN 03. So as you can see leaders here also cancels out and now we have moles of aging N. 03 that we're reacting. So now we can do a multiple comparison. Of course these are going to be easy because it's just 1 to 1 mole ratio. So one more of a G. You know, three produces 1 mold on the solid A G. B. R. Okay. So now we've essentially canceled out molds of silver nitrate. And now we're left with molds of um silver bromide. Now we can go ahead and convert the moles into mass by using the molar mass of sodium bromide. Were not given that information, but it's really easy to calculate that if we have a periodic table handy. Okay, so let's go ahead and figure this out. So we have sodium. I'm sorry. We have silver and browning. If we take a look at the periodic table, we'll see that um silver weighs one oh 7.87 g per mole and then bruning weighs about 80 g per mole. And when we add those two up, those masses up we get one .87 and that is in the g per mole. So in one more We're gonna put one mole on the bottom of a G. B. R. There are 187.87g. And that goes on the numerator. 187 0. gramps. Okay, so now we can cancel out the moles of silver bromide. And now once we do all of this multiplication and division, we're going to get a number an answer in the mass of silver bromide and that is going to be 11. grams of A G. B. R. All right. So is this our answer. Well no we we don't know yet it might be our answer. So let's go ahead and just um note that and then we're going to do the same thing. But we're going to use the other reactant. Now if we were only given information on one of the reactant and asked for what's the mass of the product? And this would be and you know that would be our answer. But because we have information on both of our reactant, we need to take a look and see which one makes the least or the most. So the second one we have for silver bromide and we have ml. So we're going to do exactly the same steps. So we're going to convert into leaders first. So we're gonna multiply it by a ratio one millimeter will go on the bottom. 10 to the negative three liters. Go on top. And now we can cancel out the middle leaders. And now we're going to multiply by the polarity of an A. B. R. Which is 2.0 molar or 2. moles per one liter. Okay, so that leaders cancel out and that's an A. B. R. Okay. And now we can go ahead and do a multiple comparison again. It's going to be just 1-1. So one more of N A. B. R. Go on the bottom of the Ratio and then one more of KGB or go on top. So now we can cancel our moles of sodium bromide. All right now, we are going to go ahead and multiply by the molar mass of silver bromide, which is 187 0. grams of KGB are per one mole of a GBR, silver blue mink. We can cancel out the moles. And now let's go ahead and do this calculation multiply and divide and we get 15.03 grams of silver bromide. Alright, so now we need to figure out which one is the answer. Well, it looks like silver nitrate. The reactant is going to be our limiting reactant because it's actually making less of silver bromide. Uh Solid hair, right? Is making that less than sodium sodium bromide wants to make. So even though sodium bromide theoretically could make more, it can because once we don't have any more silver nitrate, the reaction stops. So in theory 11.27 g of silver bromide. It's solid is all that we can make. Okay, so that is going to be our answer. 11.27 g of silver bromide. Alright, thank you so much for watching. And we'll see you in the next video