The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to convert all the iron to Fe3+ and then titrated with Sn2+ to reduce the Fe3+ to Fe2+. The balanced equation is:
What is the mass percent Fe in a 0.1875 g sample of ore if 13.28 mL of a 0.1015 M Sn2+ solution is needed to titrate the Fe3+?

Question

The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to convert all the iron to Fe3+ and then titrated with Sn2+ to reduce the Fe3+ to Fe2+. The balanced equation is: 
What is the mass percent Fe in a 0.1875 g sample of ore if 13.28 mL of a 0.1015 M Sn2+ solution is needed to titrate the Fe3+?

Accepted Answer

Hello everyone today we have the falling problem. The amount of iron and ore was determined using chroma tree where in micro mate is used to oxidize iron two plus to iron three plus. The net ionic equation is as follows. And this is balanced, calculate the mass percent of iron and a 30.2485 g sample of ore. Given that 21.41 millimeters of 0.167 moller of chrome eight or die chrome eight was used to titrate all of the iron. So the very first thing that we want to do as we stated in the question is to find the mass of iron. And so how do we start that? Well, we usually start with our given and that's going to be our 21. middle leaders. We essentially want to go from middle leaders all the way to grams the mass of iron. And so we do that through a series of steps. So the first thing you wanna do is we want to transform this into leaders so that we use the conversion factor that one. Middle leaders equal to 10 to the negative third. Leaders are units of middle leaders will cancel that. And then we multiply by our polarity of our di chrome eight similarities in terms of moles per liter. So that is 0. moles over one liter of di chrome eight. Furthermore, we then use a multiple ratio because we want to get a unit in terms of iron. So using our chemical reaction we can use the coefficients to help with our ratio. So the coefficient in front of our diet. Crow mate is one so therefore that is one mole of micro mate equals six moles of iron. We simply used the coefficient for multiple ratio. Last but not least. We have to multiply by the molar mass of iron To get g. And so to do that, we use one mole of iron is equal to 55.85 g of iron. And this is per the periodic table. Our units of liters cancel out moles, Kodachrome animals of iron cancel out. And this gives us 0. g of iron. But this is not our final answer. The second step that we have to do and this is finding the percent mass of iron is using our mass of iron that we just sold for. And dividing that by our mass of your. So it's essentially just finding out how much of our ore contains iron. And then of course because they want a percentage, we multiply by 100 And so filling in for a massive iron. We have 0.1198 g of iron. And for a massive or given In the question we have 0.2485 g of four Multiplied by 100%. And we get 48. Of iron. So 48.2% of this ore is composed of iron. And with that we've answered the question overall, I hope this helped, and until next time.

Verified Solution

Key Concepts

Redox Reactions

Titration

Mass Percent Calculation