A mixture of CuO and Cu2O with a mass of 10.50 g is reduced to give 8.66 g of pure Cu metal. What are the amounts in grams of CuO and Cu2O in the original mixture?

Video transcript

hi everyone for this problem, it reads seven g of pure metal is produced by the reduction of a 5.18 g sample containing iron three oxide and iron. Two oxide. How many grams of iron, Three oxide and iron to oxide are in the original sample. So the question that we want to answer is the amount of grams of iron, three oxide and iron to oxide in the original sample. Okay, so let's go ahead and define X as our mass Of iron to oxide and will define why as our mass of Iron three oxide. Okay, so we have those variables defined. Let's go ahead and write the molar mass for both as well because we'll need that. So the molar mass of iron of iron to oxide is 71.884 g per mole. And the molar mass of iron three oxide Is 159. g per mole. So the place that we're going to want to start. So we know we have a 5.18 g sample. So let's figure out How many moles of iron is in this. 5.18 g sample. Okay, so we have 5.18 g of iron. And we want to go from grams of iron, two moles of iron. So we need the molar mass and one mole of iron. There is 58. g of iron. With that. We see that our grams of iron cancel and we're left with moles of iron. So when we solve this, we get 0. moles of iron. So now that we know how many moles of iron we have. We can set our equation equal to this because we know that this is the total amount. So what we're going to do is we're going to say our moles of iron to oxide plus two times are moles of iron. Three oxide is equal to um moles of iron. Okay. And we have two times our moles of Iron three oxide because we have the two moles of iron there. So we're going to take we're going to multiply that by two. Okay, so we set this equal to the moles of iron that we just solved for. So now let's go ahead and plug in our variables and how we define X and y. We're going to plug that into this equation here. So for our moles of iron to oxide, we said that this is X. Okay, so this X is going to represent our mass of iron Two oxide. And we want to multiply this by the molar mass of iron to oxide. So we're going to multiply this by one mole of iron to oxide. Over .884 844. Excuse ME, of iron to oxide. Because we'll see here that the grams is going to cancel. Okay, But we don't know what exes exes are mass and that's what we're solving for. So then we're going to add two times we defined why as our mass of iron to oxide. So this is going to be two times why? And we're gonna multiply this by the molar mass. So the molar mass for iron three oxide is one mole of Iron three oxide Divided by 159.69g of Iron three oxide. We're gonna set this equal to the total moles Which we said was 0. 276 moles of iron. Okay, so now that we did that we have our equation and we know that X plus Y equals seven. And we can rearrange this so that we're solving for X. So when we rearrange this so that we're solving for X, we get X is equal to seven minus Y. So we can now plug in X. Which we have here, we can go ahead and plug in seven minus y there. Okay, so let's go ahead and do that. So we're gonna take seven minus y. Times one more of iron to oxide. Over 71.844 g of iron to oxide plus two times Y. Times one mole of iron. three oxide Divided by 159.69 g of Iron three Oxide is equal to It's going to be the same thing. We're just copying what's above 0.09- six. Let's move this over Running out of room. So we have zero 09276 moles of iron. Alright, so now we can go ahead and solve for X and Y. So let's go ahead because we substituted X. Here, we can solve for Y. First. When we solve for Y, we get Y is equal to 3. g of iron, three oxide. And we know that X is equal to seven minus Y. So we know that X is equal to seven minus 3.35. So that gives us X is equal to 3. g of iron to oxide. So let's go ahead and just write this out as our final answer. We have 3.35 g of iron, three oxide and 3.65 g of iron two oxide as our final answer. And that is it for this problem. And that is the end of this problem. I hope this was helpful