A compound with the formula XOCl2 reacts with water, yielding HCl and another acid H2XO3, which has two acidic hydrogens that react with NaOH. When 0.350 g of XOCl2 was added to 50.0 mL of water and the resultant solution was titrated, 96.1 mL of 0.1225 M NaOH was required to react with all the acid. (b) What are the atomic mass and identity of element X?

Video transcript

Hello. Everyone in this video we have two asses Hbr and H. X. 03. They're being produced when a compound with the formula X. O. To be are reacting with water. So H. X. 03 has one acidic hydrogen which is this age of course. And this reacts with K. O. H. A sample of Xo two br weighing at this many grams was added to 100 mL of water. The solution obtained was nitrate ID with K. O. H. Here in this question we're trying to determine the atomic mass and the identity of the element X. If 100 million of one molar K. O. H. Is required to neutralize all of the acid. Alright so first let's go ahead and wrap up the balanced equation for the reaction. So for my starting materials we have X. O. To be our This is a quiz and this is reacting with liquid H. 20. And this reaction then yields H. Br. As well as H. x. 0. 3. Alright first we can go ahead and determine the number of moles of K. O. H. That is required to react with all the acid. So and here just two notes our moles and moles of R K. O. H. Exactly how we can solve for this is following this equation. This is the concentration. So the polarity of R K. O. H. Times the volume of the K. O. H. If you recall this count. Em for me larry T. The units for similarity is moles per leaders. We multiply the volume that we go ahead and cancel out the units of leaders leaving us with just our moles on top. And again, we're multiplying this. Alright, so first we're gonna go ahead and actually change the units for the given amount of K O H. So we're given that we have 100 mL. We want to go ahead and convert this into leaders. So for every one leader we have one times 10 to the third milliliters, we can see here that the millions will cancel. So once I put that into my calculator, I get the value of 0.1 liters being the volume of K. O H. Now, starting from my end here, just using that equation there. So my end of K O H. So the given concentration is One moller and that's just one mole for everyone. Leader. Again, I'm multiplying this by the calculated volume that 0. L. So now we can see that the leaders will cancel, giving us the units of moles for answer. So once I put that into my calculator, I get 0.1 most of K O H. All right, so now that we determine the number of moles of K. O. H. Let's determine the moles of H. Plus that reacts with the K. O. H. So again, N stands for our moles here is something for our H plus. So we have a 0.1 moles of R K. O. H. And this reacts with or this is multiplied with. Let's see So that for every one mole of H plus we need one mole of K. O. H. We're just looking at the chemical occasion. We can see that we have a 1 to 1 ratio. So once you put that into calculator for units, why we see that the most of K O H will cancel. So for my numerical value I get 0.1 and units of course just being moles of R. H. Plus. And since there are two assets, half of the total moles of our H plus will come from HBR and the other half will come from our H. X +03. And there's one mole of H plus per mole of H X +03. So the moles of our H X 03 is equal to 0.1 moles of our H plus divided by two Plus one mole of our HX 03 over one mole of R H plus. So once you put that into the calculator we get the answer to be 0.5 moles of our H X. +03. So based on the balance occasion above, there's one mole of H XO three produced per one mole of X +02, br go ahead and scroll down low for more space here. Alright, so Now I can go ahead and software the mold. So end of the XO to be our starting off with 0.05 moles of HX 03. We know that for every one mole of the X. Oh to be our mini one mole of HX 03, you'll see here that the moles of H R H X 03 will cancel. So once you put this in your calculator we get the value to be 0. moles of H R. X. O. To be are now we can go ahead and determine the molecular weights of this. X. 02 Br Let me go ahead and scroll down for this. Alright, so molecular weight is equal to M for the mass over N. Which is the moles of course were regarding our X. 02 br. All right, so the mass of the XO to be our is 6.295 g. So, let's go ahead and put all this into the equation again. The M for mass is equal to 6.295 g. The N is 0.205 moles. Once you put that into a calculator, we get the molecular weight to be 1 25.9 g per mole. So very no the atomic weight of auction and grooming. So let's actually write this out. So for our auction this is 16 g per mole. And for our booming B. R. This is 79.9 g per mole. So let's go ahead and subtract it just to find out what X is exactly. So taking the overall molecular weight of 1 to 5.9 g per mole, we need to subtract this. Well, let's see here for our creation X. O. To be are we know that we have two atoms of auction. So that's two. And then for each oxygen this contributes 16 g per mole. And then subtracting are booming. We just have one. So we subtract 79.9 g promote. So everything is in the same unit. So we can just simply subtract this. I'll go ahead and put this into my calculator. Once I do, I get the numerical value to be 14. g per mole. Again, this is the atomic weight Of X. So that's one of the answers for our question. And just taking a look at the pier a table. What has a mass of 14 g from all. And this is going to be nitrogen of course, with a symbol of just end on a pure a table. So this is going to be my final and last answer for our question