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Fe can produce Fe2O3 in accordance withthis reaction equation:Fe2O3(s) + 3CO(g) → 2Fe(s) +3CO2(g)167 g Fe2O3 reacts with 85.5 g CO andforms 72,3 g Fe. Calcul...

Question

Fe can produce Fe2O3 in accordance withthis reaction equation:Fe2O3(s) + 3CO(g) → 2Fe(s) +3CO2(g)167 g Fe2O3 reacts with 85.5 g CO andforms 72,3 g Fe. Calculate how many grams Fe can be formedand the percentage-wise yield?Given Molar masses:MC = 12,01 g/molMO = 16,00 g/molMFe = 55,85 g/mol

Fe can produce Fe2O3 in accordance with this reaction equation: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 167 g Fe2O3 reacts with 85.5 g CO and forms 72,3 g Fe. Calculate how many grams Fe can be formed and the percentage-wise yield? Given Molar masses: MC = 12,01 g/mol MO = 16,00 g/mol MFe = 55,85 g/mol



Answers

Iron(III) oxide reacts with carbon monoxide according to the equation:
Fe2O3(s) + 3 CO(g)-2 Fe(s) + 3 CO2(g)

A reaction mixture initially contains 22.55 g Fe2O3 and 14.78 g CO. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains?

Here we have iron oxide reacting with carbon monoxide to produce iron, metal and carbon dioxide according to this balanced chemical reaction. If a reaction mixture initially contains 22.5 g iron three oxide and 14.78 g carbon monoxide, we should be able to determine which is the limiting reactant, how much product is created and the excess of the reactant that is not limiting. It is this last question that they are asking you to answer for this problem. So the first thing we need to do is identify which is the limiting reactant. By calculating the amount of any product you choose that can be created by each reactant. The reactant that produces the least amount of product will be the limiting reactant and the other one will be in excess where the limiting reactant is consumed completely. So if we had 22.55 g of iron three oxide, we can convert those grams to moles by dividing by the molar mass and then convert the moles of iron three oxide to I'm going to choose iron as the product and we would get 0.2824 moles. Iron created if all of the carbon monoxide were consumed, will convert the 14.78 g carbon monoxide into moles by dividing by its smaller mass. Then we'll go from moles of carbon monoxide, two moles of iron and we get 20.3518 moles. So the one that produces the least amount of product is the limiting reactant. The other one that being iron three oxide. The other one carbon monoxide is in excess. So the maximum amount of product that we're going to create is the 0.2824 moles of iron. So let's now calculate how many grams of carbon monoxide are required to produce the 0.2824 moles of iron will go from moles iron two moles carbon monoxide with the 2 to 3 relationship, and then molds carbon monoxide, two g carbon monoxide, 11.87 g carbon monoxide are consumed. So we'll subtract that from what we start with 14.78 g to get 2.91 g carbon monoxide remaining.

So let's look at the reaction of Iron three oxide with carbon monoxide to make iron and carbon dioxide. And Let's say that we started out with one kg Of Iron three Oxide. And in the process of doing this reaction, uh, say that we come up with, Uh, of of iron that was made. And uh, we're gonna wind up looking for the percent yield for this particular reaction. So, let's do some strike geometry to figure out how much iron we should have made. And then convert that um into percent yield. So when we're doing the math and using molar mass in grams per mole, we can't use kilograms, we're going to have to use grams. There's 1000 g in a kilogram. So we have 1000 g of iron three oxide. And we'll use the molar mass of Iron three oxide To convert two moles, Which is 159.69 g. To convert to iron, will use the mole ratio. We make two moles of iron For every one mole of iron. Three oxide. And then we'll convert two g of iron by using its molar mass 55.845 g per mole. Yeah. Okay. and 2 3 significant figures Will get 600 99 grams of iron. This is the amount that we we theoretically could make if every one of these two molecules reacted. But it turns out that they didn't completely react. Only 654g was made. So we will determine the percent yield by taking the actual yield. And that is determined in the lab divided by our theoretical yield, which is determined by the math and multiplied by 100. So in the lab We made 654 g. Yeah. You just make that a four. Yeah. And in theory the math says we could have made 699 g to get a percent. We multiply by 100 We got 93.5% yield.

This problem is also a limiting reactant problem where we are given to amounts for two different reactions and we need to figure out which one is the limiting reactant and then from the limiting reactant to determine the theoretical yield and ultimately the percent yield. The balanced chemical reaction is two moles of iron. Three oxide reacts with three moles of carbon to produce four moles iron And three most carbon dioxide. So if one of the reactant is iron, three oxide and we have 12.5 g of it, we can convert those grams into molds by divided by the molar mass of iron three oxide than knowing the moles of Iron three oxide required to produce four moles of iron. This being the ST geometry, we can calculate the molds of iron produced, then multiply by the molar mass to convert the moles iron into grams iron and we get 8.74 g iron produced from 12.5 g iron three oxide. Next, we'll do the same thing with carbon. We have 2.6 g of carbon, which we can convert to moles carbon by dividing by the molar mass carbon. Then, when we know the moles carbon, we can calculate the moles of iron With the 3- four strike geometric relationship. Then when we have moles, iron will multiply by the molar mass iron to get graham's iron And if all 2, 6 g of carbon were consumed would get 16 g of iron. Therefore, iron three oxide produces the least amount of iron. This being the theoretical yield 8.74 g and iron three oxide producing the least amount of iron is the limiting reactant Percent yield will be what we actually get, which is provided in the problem, divided by the theoretical multiplied by 100 and we get 80.1 yield.


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