On heating a 0.200 g sample of a certain semimetal M in air, the ...

Question

On heating a 0.200 g sample of a certain semimetal M in air, the corresponding oxide M2O3 was obtained. When the oxide was dissolved in aqueous acid and titrated with KMnO4, 10.7 mL of 0.100 M MnO4- was required for complete reac- tion. The unbalanced equation is X (b) How many moles of oxide were formed, and how many moles of semimetal were in the initial 0.200 g sample?

Accepted Answer

Hello. In this problem we are told a 0.350 g sample of a certain semi metal m needed an air yielded and matching oxide reaction needed 7.67 mL of a 0.125 miller die cremate solution when the oxide was dissolved in aqueous acid and tie traded with potassium bromate. The unbalanced equation for the tight relation of our semi metal tri hydroxide is provided and we're told this takes place under acidic conditions. Were asked how many moles of semi metal were present in the initial 0.350 g sample and how many moles of the oxide were formed. To begin. Then we will balance our reaction. We'll start with the reduction half reaction. So we have then chromium being reduced from an ox stage data plus six in micro mate to unlock stage data plus three. We need to begin by balancing those things other than hydrogen oxygen. So we balance our chromium by adding a two to the product side. But then got to balance our oxygen's by adding water. So we need seven waters to balance the oxygen and then we'll balance the hydrogen by adding hydrogen ions. So we have 14 on the product side. So we need 14 hydra nines on the reactant side. And then we're gonna balance our charge. We have a net charge of 12 plus on the reactant side and plus six on the product side. So we'll add six electrons to the reactive side. To get our charge to balance. Then we'll balance the oxidation half reaction where we have our some metal tri hydroxide. There's something about being oxidized to form a some metal tri hydroxy oxide. We begin by balancing all those things besides hydrogen oxygen. So at this point are seven metal is balanced. So we balance the oxygen by adding water and then we balance the hydrogen by adding hydrogen lines. We have five on the reactive side, so we need two more on the product side to get our hae jin's to balance and then we balance our charge, it's neutral on the reacting side. And so to get it neutral on the product side, we add two electrons. We then can combine these two have reactions to come up with our overall reaction in order to do that. Then the number of electrons being gained has double the number of electrons being lost. So we'll multiply our oxidation half reaction through by three. In doing so then we have six electrons being gained and six being lost. So those will cancel. We also see that we have then three times two is six hydro nines on the product side, 14 on the reacting side. So we can get six of those to cancel. And we'll be left with eight hydro nines on the reacting side. We have three waters on the reacting side and seven on the product side. So we can cancel three of those waters and be left with four on the product side, combining what's left over. We then get potassium micro mate Reacting with three moles of our semi metal try hydroxide and eight hydro knin's Going to form two moles of chromium, three Plus three moles of our semi metal tri hydroxy oxide And four moles of water. We can then make use of this balanced equation to determine the moles of semi metal originally present. So we are told that takes 7.67 ml object, die pro mate than to titrate our some metal tri hydroxide will convert them from the Leaders. Leaders make use of the concentration of the That girl made solution which is .125 moles every leader and then looking at our balanced reaction for every one mole of micro mate that reacts. We have three moles of our summer metal tri hydroxide. So our milliliters of di chrome cancels. Leaders cancels moles of di chrome cancels and we're left within 0.288 moles of our semi metal try hydroxide. You can then make use of this quantity of our some of that will try hydroxide. Find the amount of our semi metal that was originally present based on the chemical formula one mole of our summer metal tri hydroxide as one mole of our semi metal. Most of our semi metal tri hydroxide cancels. And this works out to 2.88 Times 10 to the - molds of our semi metal that was originally present. We then make use of are moles of our semi metal to determine the moles of are some of the metal oxide that were formed based on the chemical formula that were provided. There are two moles of semi metal in every one mole of our send a metal oxide, the molds of our summer metal cancel. This works out then to 1.4, 4 Times 10 to the -3 moles of our talk side, the small amount of our semi metal and are oxide and corresponds to answer D. Thanks for watching. Hope. This helped.

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