Pure oxygen was first made by heating mercury(II) oxide: HgO --> (heat) Hg + O2 Unbalanced (a) Balance the equation. (b) How many grams of mercury and how many grams of oxygen are formed from 45.5 g of HgO? (c) How many grams of HgO would you need to obtain 33.3 g of O2

Video transcript

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.