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Ch.3 - Mass Relationships in Chemical Reactions

Chapter 3, Problem 118

Element X, a member of group 5A, forms two chlorides, XCl3 and XCl5. Reaction of an excess of Cl2 with 8.729 g of XCl3 yields 13.233 g of XCl5. What is the atomic weight and the identity of the element X?

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Hello. Everyone in this video we have a reaction with Y cl two reacting with cl two. And this forms Y cl four. So we need to go ahead and identify what element Y. Is. How we can do this is by calculating for the molar mass of Y and comparing it to all the molar mass is present in our peer at the table. But of course we need to first calculate for the grams of element Y. As well as the molds of element Y. And here we can start off by simply writing the chemical reaction that occurs. That's when Y cl two reacts with excess cl tl to yield Y C L four are only products here. So first we go ahead and calculate for the mass of chlorine that reacts and that's just using the two values present in our given problem. So that's 19.13g. Subtracting by 11.33 g. And this gives us 7.80 g of chlorine. So that is how much of chlorine lies reacted in this chemical reaction. So the second part now is we can go ahead and let X equal to the mass of chlorine in our products because that's what we're interested in as of now. So to calculate for the mass of why that's just taking the 19.13 g of the product. Subtracting that with X. Which is the massive flooring inner product. And that gives us the mass of wipe. All right. So in our products we have our X. Which we said to be the mass of chlorine and Y. Sale for. So for every one X we need four chlorine. And that's just the ratio I got from the product. And then we just calculate for The mass of chlorine that's reacted. So that's 7.80 g and that's just for every two clones. So that's just the reaction that we got from our starting materials. So to isolate X. We can do that by Multiplying both sides by four chlorine, sergeant nominator. Here you see that the four corners will go ahead and cancel. So that goes ahead to isolate our ex. I can see what we put in. The calculator is just 7.80 was applied by four divided by two. Alright, So X is going to be equal to 15. or 2.6 I apologize 15. g. And this is the mass of chlorine. So now that we have solved for X we can go ahead and use this equation here that we wrote earlier. So 19.13 g minus X. Which we solve for to be 15.6 g. And now give us the mass of Y. So the difference here is 3.53 g again, this is the mass of Y. Which is going to be our numerator for our molar mass. So now we can go ahead and solve for the denominator which is going to be the moles of Y. So we can go ahead and start off converting our grams of chlorine, which is our X into moles. And then we use a multiple ratio to cover into molds of why are unknown. So again, we're starting off with the mass of chlorine that reacts which is 15.6 g. We can go ahead. Use the molar mass of chlorine to convert our ground declaring into moles of chlorine. So for every one mole of chlorine we have 35.45-7 g of chlorine. You can see here that the grounds of chlorine will cancel. And that gives us a value of 0. moles of chlorine. Now that we have the moles of chlorine. Like I said, we're gonna go ahead and use the multiple ratio. We can see from a product Y cl four that for every one mole of y we will need four moles of chlorine. So that's going to be my conversion factor. That will go ahead and use again, we're starting off with the 0. moles of chlorine. We're gonna go ahead and use that multiple ratio to convert the moles of chlorine into our moles of white. So we see here that the moles of chlorine will cancel. And that gives us a value of 0.11 moles for why now solving for the molar mass of Why now that we have both components. So we need grams or mass on top, which is the three 53g. And then we need the moles which is 0.11 moles. So putting that into my calculator, I get out the value of 32.0909 units being g per mole. So now, using that value that we just calculate for and looking at the pr table, sulfur will have the closest solar mass to this calculated molar mass. So then we can conclude that element, why is equal to sulfur? And this right here is going to be my final answer for this problem. Thank you so much for watching.
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