(c) What is the identity of the semimetal M?

Video transcript

Welcome back everyone a 0.175 g sample of a specific semi metal m heated and air yielded the corresponding oxide M. O. To the reaction required 17.50 mL of 0. moller permanganate an ion when the oxide was dissolved in aqueous acid and tie traded with potassium permanganate. So we are given an unbalanced reaction here and we need to give the name of the semi metal M. So we're going to begin by writing out half reactions based on our reactant is given in the overall reaction from the prompt. So our first reaction is H two M 03, which will assume is our Aquarius acid. And as a product it's going to form H two M 04 which is an Aquarius product. For our second half reaction, we have our permanganate an ion which as a product is going to form our manganese two plus kati on. So our first step is to balance out our atoms in each of our reactions. We can see that we have four moles of oxygen on the product side, but only three moles of oxygen on the reactant side. And we want, we want to recall that we balance oxygen using water. So we're going to expand our react inside and add water as a reactant, specifically just one mole of water. So adding one mole of water bounces out our oxygen's on both sides. However, now we have a total of four moles of hydrogen on the reactant side and only two on the product side. So next we want to recall that we balance hydrogen Using hydride, meaning that on our product side we're going to add two moles of hydride and this is now going to give us four moles of hydrogen on the product side and four on the reactant side. So now our atoms are balanced for our first half reaction. And now let's go and look at our second half reaction where we see, we have four moles of oxygen on the reactant side and none on the product side. So we're going to need to add four moles of water on the product side for the second half reaction. And now we've introduced hydrogen where we have eight moles of hydrogen on the product side. So we'll need to expand our reactant side And add eight moles of hydride on the reactant side. Now that we have our atoms balanced here, we need to balance out the charges in both half reactions. In our first reaction, we have a net charge of plus two, which we can balance out on both sides by canceling that plus two charge by adding two electrons on the product side here. And now for our second reaction, we have a net charge on the we act inside of zero because this plus one cancels out with the minus one but on the product side we have a net charge of plus two. And so we're going to need to or sorry correction, we have a coefficient of eight on our reactant side in front of the plus one charge. So we have a net charge of eight minus one, meaning we have a net charge of plus seven on the reactant side. However, we have a net charge of plus two on the product side. So we're going to need to add electrons to are reacting side to get that plus seven charge down to minus or plus two rather meaning we're going to add five electrons to our react inside here. Now we need to come up with an overall reaction by bouncing out the electrons in both reactions. So we'll be able to take this entire first reaction and we want to multiply everything by a factor of five. And then for the second half reaction to get our electrons to balance out with the first one, we're going to multiply everything by a factor of two. And so what we're going to have is and actually I'll use a different color for the second reaction, let's multiply everything here in purple by a factor of two. So for our first half reaction, we're going to Now, once we have everything multiplied by five b five moles of H two M 03 Plus five moles of water produces five moles of H two M Plus 10 moles of hydride Plus 10 moles or 10 electrons. And now for our second half reaction, we're going to multiply everything by a factor of two, which is going to give us two moles of permanganate an ion. And we need the AQ symbol here. This is then added to 16 moles of hydride Plus 10 electrons and produces two moles of our manganese, Catalon And eight moles of water. So now we're going to add these two half reactions together now that our electrons are balanced And we're going to come up with an overall equation where before we do so we want to cancel out what's similar. So we can get rid of the electrons here on the product side and here on the react inside. We can cancel out 10 of the 16 hydride with the here on the product side. That would leave us with six hydroids. For our second reaction, we can cancel out five of the waters with the eight here on the product side, which would leave us with three moles of water for our second reaction. And now we're just gonna carry down everything that we're left with. So that we have an overall reaction of five moles of H two M 03 Plus two moles of permanganate an ion yields, or sorry, plus or six moles of hydride that we were left over with Yields two moles of our manganese Catalon, Plus five moles of Rh two M And then plus three moles of water. And this is a liquid symbol here. And so we're going to use the coefficients from this overall reaction to find the moles of our reactant which will ultimately lead us into the moles of our metal in the metal oxide, which as the prompt states is dissolved in our acidic acquis react ints. And again tie traded with permanganate, potassium permanganate. So we're going to begin by noting down the volume of the solution given in the prompt as 17.5 ml. But we want this to be in leaders. Since we're given the concentration in the prompt of Armando an ion of our permanganate an ion in leaders. We're going to multiply this to convert to leaders. So we have male leaders in the denominator and leaders in the numerator recall that our prefix milli tells us we have 10 to the negative three of our base unit. Leaders canceling out middle leaders were left with leaders and we'll have our volume of our solution equal to 0.175 leaders. Now using this volume, we want to find moles of our permanganate an ion so mn 04 minus And we would take that volume of our solution .175 L and multiply it by our polarity given in the prompt for permanganate an ion as .550 moller which were interpreting, we recall as moles per liter of permanganate. And so this is going to allow us to cancel out leaders leaving us with moles of permanganate Ion. And this is equal to a value of 9. times 10 to the negative third power moles of our permanganate an ion. So now that we have moles of our permanganate, we want to find moles of our acidic acquis reactant given in our prompt as H two M. 03. And this is what our metal is dissolved in. So We're going to use our moles of permanganate, an eye on that we just found 9.6-5 times 10 to the negative third power moles of permanganate. And we're going to multiply by a molar ratio between moles of our permanganate, an ion two moles of our H two M 03 reactant, which we'll get from our overall balanced reaction where we see we have a 2 to 5 molar ratio. So we have two moles of our permanganate. An ion for five moles of our H two M. 03. Acquis acidic reactant. So canceling out moles of permanganate an ion, we're left with moles of our acidic Aquarius reactant and we're going to have an amount equal to 0. 406 moles of our H two M 03. Again, our acid used to dissolve the metal. Now we want to find our moles of our metal oxide, M203. Or sorry, that is defined in the prompt as M. 03. And so this is set equal to our moles of our acid. Used to dissolve our metal oxide that we just found being 0.02406 moles of H two M 03. We're multiplying this by the ratio between moles of our H two M acid and molds of our metal oxide, M. 02, which in the formula we can get that we have a 1 to molar ratio to get M 02 out of H two M 03, which allows us to cancel out moles of H two M 03. Leaving us with moles of our Metal oxide equal to 0.02406 moles of M. O. To our metal oxide. And now that we have moles of our metal oxide, we want to find moles of our semi metal em in our metal oxide M. 02. And so we would take our calculation from before 0.2406 moles of our metal oxide, M 02. And multiply by ratio between moles of our metal oxide, M 02 and moles of our semi metal M. Which we can see is again a 1 to 1 molar ratio. When we look at the formula for our metal oxide, we have one mole of em. And so we can cancel out our moles of metal oxide leaving us with molds of our semi metal M equal to 0.2406 moles of our semi metal M. And now that we have molds of our semi metal, we can find its mass to determine its identity using our periodic tables. So we want to use the mass of our sample given in the prompt that is used for the semi metal given as 1. g. And we're going to divide this by our molds of our semi metal that we just found being 0.2406 moles of our semi metal M. This is going to equal a value of 72.73 g per mole as our units, which we can see is a molar mass unit. And so we want to go to our periodic table And we'll see that this is close to about 72. g per mole, which is our molar mass of Germanium, our Adam G.E. and so we can say that therefore the semi metal is Germania and germanium would be the final answer. To complete this example as the name of our semi metal M. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video