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University of Maine
Problem 140

High-temperature superconducting oxides hold great promise in the utility, transportation, and computer industries
(a) One superconductor is $\mathrm{La}_{2-x} \mathrm{Sr}_{x} \mathrm{CuO}_{4}$ . Calculate the molar
masses of this oxide when $x=0, x=1,$ and $x=0.163$
(b) Another common superconducting oxide is made by heating a mixture of barium carbonate, copper(II) oxide, and ytrium(III) oxide, followed by further heating in $\mathrm{O}_{2}$ :
4 \mathrm{BaCO}_{3}(s)+6 \mathrm{CuO}(s)+\mathrm{Y}_{2} \mathrm{O}_{3}(s) \longrightarrow
2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{65}(s)+4 \mathrm{CO}_{2}(g)
2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{6.5}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7}(s)
When equal masses of the three reactants are heated, which reactant is limiting?
(c) After the product in part (b) is removed, what is the mass $\%$ of each reactant in the remaining solid mixture?


a) $x=0 \rightarrow \mathrm{La}_{2} \mathrm{CuO}_{4}: 405.37 \mathrm{g} / \mathrm{mol}$
$x=1 \rightarrow \mathrm{LaSrCuO}_{4}: 354.08 \mathrm{g} / \mathrm{mol}$
$x=0.163 \rightarrow \mathrm{La}_{1.837} \mathrm{Sr}_{0.163} \mathrm{CuO}_{4}: 397.01 \mathrm{g} / \mathrm{mol}$
b) limiting reactant: $\mathrm{BaCO}_{3}$
c) $0 \% \mathrm{BaCO}_{3,} 60.5 \% \mathrm{CuO}, 28.6 \% \mathrm{Y}_{2} \mathrm{O}_{3}$



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Video Transcript

given the following general formula for a semiconductor, we can find different formulas as well as their molar masses simply by substituting in for the variable. So if, for example, x zero, that gives us a compound L A to see you, 04 And we find the molar mass simply by looking up each of the different elements on the periodic table and adding them together. That's to L. A, says one copper and four oxygen's. If X equals one, our formula ISAS follows, and the molar masses again found by adding one land them once drawn him one copper and four oxygen's Sometimes in semiconductors, we see that we have not whole numbers, and we just substitute again, as we did before, and find our Mueller Mass exactly the same, except for this time multiplying the molar mass of Lampton um, by 1.837 and strong, and by 0.163 for molar mass of 300 and 97 grams per mole. Second part of the question gives us a reaction between barium carbonate and copper oxide in the atrium compound. Do you make a semiconductor with the falling formula as well as carbon dioxide. And the question asks if we have equal amounts of each substance, which will be the limiting re agent. So it's helpful to know the molar mass of each before we start. So if we have equal masses, we could either use X or in this case, I'm just going to assume we have 100 grams of each to find the limiting re agent. We're looking for the starting material that produces the smallest amount of the product. So I changed each two moles using the molar Mass. And then I find the moles of product using the coefficients from the balanced equation. And I do this for each substance and whichever produces the smallest amount of product is the limiting re agent. Well, if I start with 100 grams herbarium carbonate, it changed two moles by dividing by the molar mass and then I multiplied by the mole ratio. There are two moles of product for every formals of barium carbonate. I do the same for my second substance, copper oxide dividing by the molar mass and thing that there are two moles of product for every six moles of copper oxide or 0.49 moles of product and finally, for my atrium oxide, changing it two moles and I see that there are two moles of product for every one mole of why 203 miss produces point 886 moles product. Because barium carbonate produces the smallest quantity of product, this is the limiting re agent we should expect to produce 0.253 moles a product. When we look at how much of the other compounds will remain, we can use this amount of product to predict how much we wouldn't need to produce. So I know that all of the barium carbonate is used up. So that means that there will be nothing left. 0% remains. To find out how much copper oxide remains, we find out how much is needed to form 0.253 moles of product. So again we start with their moles and then changed to moles of copper oxide using the balanced equation. There are six moles. See you. Oh, for every two moles of product. And then I changed that to mass by multiplying by the molar mass. So any aid 60.4 grams. But I started with the 100 grams. So that means I have 39.6 grams. Remain same thing for y 203 If I want to make 2.53 moles of product from the equation, I see that there are two moles of product for every one mole. Why 203 and I multiply by the molar mass so that I see I need 28.6 grams because I started with the 100 grams. I have 71.4 grams remaining for the percentage copper oxide is what I have remaining divided by the total mass of the two products. Times 100 or 35.7%. My percent. My math. Why 203 is the mass remaining divided by the total mass Of the two products times 100 we're 64.3%

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