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Is used to titrate a sample containing an Potassium dichromate The sample is dissolved in unknown percentage of iron: of the iron to Fez* ions. The solution HzPos/H...

Question

Is used to titrate a sample containing an Potassium dichromate The sample is dissolved in unknown percentage of iron: of the iron to Fez* ions. The solution HzPos/HzSO mixture to reduce all 0.01500 M K,CrzOz producing Fe:t and Cre* ions in is then titrated with

is used to titrate a sample containing an Potassium dichromate The sample is dissolved in unknown percentage of iron: of the iron to Fez* ions. The solution HzPos/HzSO mixture to reduce all 0.01500 M K,CrzOz producing Fe:t and Cre* ions in is then titrated with



Answers

Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 $\mathrm {mL}$ of 0.0100 $M$ $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is required in the titration. What percentage of the ore sample was iron?

Calculate the mass percent of iron in hemoglobin will first need to Create our balanced equation. MN 04-. To amend two plus would be one half reaction and F E two plus two. F E three plus would be the second half reaction stones. These half reactions separately. Mhm And multiply its one x 5. Yeah. Trans cancel. Mhm. Mhm. Mhm. Yeah. Mhm. Mhm. And here is our balance occasion. First calculate the malls of the potassium permanganate from the given more clarity and volume. The polarity is zero 2100 moller for you Is 32.3 ml or .03- three L. This gives us point 0000 678 moles of the K M n 04 minus. Now let's use our reaction study geometry to find a mass of iron. Yeah, one mole of cam in '04 To one mole of MN 04 one mole of M n 04 minus from our balanced equation, it's equivalent for five malls, especially two plus one more FB two plus to one more iron. one mole of iron, The mouse at 55.8 g and this is equal to .0189 g of iron. No, we can find our mass percent of iron by taking the massive iron that we just calculated dividing this by the mass of the hemoglobin. Multiplying by 100 Mass of the iron was zero 189 grounds. The mass of the hemoglobin Is 5.00 g. Therefore the mass percent of iron and hemoglobin works at 2.378%. Mhm.

So what we're gonna do with this problem is we're going to take the polarity and the volume of the potassium permanganate and use that to figure out the amount of iron that it reacted with. And then once we have that massive iron will be able to find our percentage. So we'll go ahead and start with the more clarity of our potassium permanganate in the volume. Don't forget to put your volume in leaders. Okay, so it's gonna show us that we needed to use a very small amount of moles of M N 04 minus. Go ahead. And we'll do our mole ratio and we'll convert permanganate to iron. We use ft two plus, because that's what everything was. The iron was converted to originally, and that's a 5-1 ratio. And then we'll go ahead and change moles to grams. We use a smaller massive iron, even though it's an eye on a couple of electrons here, there aren't going to make any difference. So this should give us a mass of, of iron. Okay, Happy to plus, or it'll be the same for the F So that's the amount of fee that must have been there. So now we're gonna go ahead and use that to figure out our percentage, So go ahead and percentages. Just our part overhaul With no 189g of our iron In our five g sample, And we'll get that. It was .379 percent iron.

Helping them. 97. We go see Miller away. Lie last. Drop them 94. But we have a couple of the mask off the ferrous ion Participate in read obstruction because they're always enough beautified that's in verify. So in the only a part of part apart amount off the or is iron So we have toe lie said before when we see solution and received a mask we see the volume and with similarity we should calculate the mom fuss And were you that's more Toto in the more radio And I have end K two AM and 04 And in this equation we see the more magnate he would among the more the comeback the Obama Mangane, Potassium about So it is he equal Pay squat, right? And I got the harmony. Three boy 73 mil Is he right? I know, But boy 23 How was it? We have boy 0 23 30 on later we have a boy 0194 m And when Conte we were 5.9 and me cover like this Oh, we have we have. Or if I do, boy, If I to 4.52 Tom, No default mark. And now we write equation with my ferrous iron. Plus that I roll my eyes This leader oxidizing agent I'm less prothom Ecologists happen in an acidic environment. And we have here boy for both. For night too. Time the 10 94 Ma on we see in this more radio sit more off iron do you? Quite only one more day roommate. And now therefore boy five proble that grow may mechanistic time for this amount And now dizzy the mall iron And now what? We know the more pie and we can find a muscle Bayan obviously buy in this equation Marcy, go more time for the monem Us Okay? And we go, huh? Rally boy 151 from iron And after with a crime My angry, confined Perceval the iron in all there are involved. If I so we've been five Send up the iron you are now 50 boy, 0%. Could the lumber boy pity for by one Almost it Good. Now with

With redox chemistry where we have the reduction and oxidation of elements within the same reaction. Where reduction is the gain of electrons, oxidation is the loss of electrons. So fastly we have the oxidation half. So so that is F E two plus two. F E three plus add an electron. So you can see that iron is further oxidized. We can look at the reduction half style, which is M N. O. For minus At eight h. Plus, add five electrons. This gives us MN two plus and fall out of water. So we can look at the overall reaction, which is M N. O for minus. Had five F E two plus At eight h plus gives those five F E three plus at MN two plus Had four lots of water. H two. Well. And so we can calculate the owner of the cell where energy is equal to e reduction at E oxidation. So that is equal to Positive 1.512 volts. I'd Negative not 769V. We have positive not .743V. And so we can continue on to calculate the malls of permanganate iron over on the next page. So we've already seen the balanced equation. So the molar ratio between the permanganate and iron two plus is 1 to 5. So we can calculate the number of moles of potassium permanganate needed required for this titillation by multiplying its polarity of not point to not point not to mola with the volume 55.63 mil. So what we get for moles of K. M and oh four, not point not to mola Multiplied by 55.63 mm. We need that in units of liters. What we get is not point not not 111 moles. So finally we can calculate the malls of iron two irons from the malls of permanganate iron because the molar ratio we are aware of. So what we do is we take Not point not not 111 models multiplied by five moles of f E two plus, divided by one mole of M n 04 minus which gives us not point not not 555 moles four f E two plus. And so finally we can calculate the massive iron in the sample by multiplying the number of malls with its atomic mass over on the next page. So what we have is not point not not 555 moles Multiplied by 55.845 g per mole To give us not .30994 g. And so we can calculate the mass percent of iron. So we have the percent of I n f E. Is equal to the value, Not .301g 310g, Divided by the mass of the sample, which is not .3500 g. multiplied by 100 to get in percentage are g. Council we get 88.6%.


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