A procedure for determining the amount of iron in a sample is to convert the iron to Fe2+ and then titrate it with a solu-tion of Ce(NH4)2(NO3)6: What is the mass percent of iron in a sample if 1.2284 g of the sample requires 54.91 mL of 0.1018 M Ce(NH4)2(NO3)6 for complete reaction?

Video transcript

Hi everyone. I have a problem giving us the following that ionic equation and it tells us that the amount of iodine three minus ions can be determined by the eye line three minus solution with known concentration of S two minus Aquarius at the sulfate ion. And asking us to calculate the similarity of an iodine three minus solution. Given that it requires 35.8 mL of 0.54 moller N A two S 203. Acquiesce to type four millimeter sample of iodine three minus a quiz. So first let's see what happens when in a two S 203 disassociate. So we're going to get to sodium plus plus one S two minus. And our is going to equal moles over leaders. So we're going to just to get The moles of S203 and then we can use that together as to the malls Of iodine 3 -. So we have zero 0.45 four malls of S her one leader. And we're going to multiply that by 35.8 million L. And that has to be changed to leaders. So we're going to multiply that by 10 to the negative third. Leaders over one m. And that gives us 0.01625 moles of S 203 two minus. Now we want to convert this two moles of red, I'm three minus. So we have our 0. moles Of 2032 -. And then we're gonna multiply that at multiple ratio. So for every one mole of iodine three months we have two moles of s two 032. And that's found on our balanced equation. So that's going to equal 0.0081, 2, 66 malls of iodine a minus. Now, we need to calculate the actual more clarity. So we have our 0. balls of iodine three minus over our leaders. And it gave us 34 ml. So we're just going to divide that by 1000. So to change it to leaders. So 0.034 l And that equals 0.239 Smalling iodine 3 -. And that is our final answer. Thank you for watching. Bye.