The YBa2Cu3O7 superconductor can be synthesized by the
sol–gel m...

Question

The YBa2Cu3O7 superconductor can be synthesized by the 
sol–gel method from a stoichiometric mixture of metal ethoxides 
followed by heating in oxygen. How many grams of 
Y1OCH2CH323 and how many grams of Ba1OCH2CH322 
are required to react with 75.4 g of Cu1OCH2CH322 and 
an excess of water? Assuming a 100% yield, how many 
grams of YBa2Cu3O7 are obtained?

Accepted Answer

Hi everyone here, we have a question telling us that the compound zirconium silicate, Z. R. S. I. 04 solid can be prepared by reacting zirconium dioxide solid and silicon dioxide solid at high temperatures. And our goal here is to calculate the molds a mass of zirconium dioxide needed to produce 3.500 g of zirconium silicate solid, assume that the reaction goes to completion. So first we're going to ride out our balanced equation. So we have zirconium dioxide solid plus silicon dioxide solid forms zirconium silica cape solid. And that is already balanced that we can move on. So our molar mass of zirconium silicate equals 91 0. g per mole for zirconium plus 28 0. g per mole for silicon plus 16 grams per mole Times four for oxygen. And that equals 183. g per mole. And now we need to calculate our molds of zirconium dioxide. So we have 3. g of zirconium silicate times one mole of zirconium silicate divided by its molar mass. So 183. g of zirconium silicate times one mole of zirconium dioxide divided by one mole zirconium silicate. And our grams of zirconium silicate are going to cancel out. And our moles of coney um silicate are going to cancel out. And that is going to give us 0. moles of zirconium dioxide. So that is our answer for the first part. Now we need to calculate the mass of zirconium dioxide needed. So our molar mass of zirconium dioxide Equals 91.2 - four g per mole Plus 16 g per mole Times two. And that's for the oxygen. And that equals 123. grams per mole. And now to calculate the mass of zirconium dioxide, we are going to take our moles, Which is 0.01909. And we're going to multiply by our Molar mass, which is 123.2 - four g. And that's going to be over one mole because it is g per mole. And our moles here are going to cancel out. And that is going to equal 2. g of zirconium dioxide. And that is our final answer. Thank you for watching. Bye.

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