Question

briefly discuss Redox discuss reaction performed discuss how moles of iron determined from titration discuss how (and report) grams of iron present and % iron in sample Discuss possible errors/improvements to experiment. Proper grammar and use of English.

Answer 1

I can't determine the % of Fe, because according to your given Molarity and Volume of KMnO_​​​4, the Mass of Fe in all the 3samples are coming out to be more than 1g. I've explained my calculations in detail, in sample 1, a 12.5mL of 0.5M MnO_{​​​​4}​​​- Solution can't contain 1.88x10^-4 moles, there must be something wrong in your given data.

HCl must not be employed in this Titration.

Reduced Stecies oxidised Species wum X cauty * REDOX Reaction Means simultaneous Roduction and Oxidation Reactions, Am Oxidant and a Reductant react to he Oxidised and Roduced Species. > Mno, It sta 24 + 8HT = Mont fe + 4H0 This is the Reaction, mentioned Oxidant Reductant - in the question. & Molarity of Mal = 0 sm) at Valoma & Modlarity - Males of KMO U sample-1 Sample 2 Sample - 30 Moles of (12.5x0.5 0.5) - 1126*0*5). . I Mole of KMnO KMnO 60 25x10 moles ²55x10mles =5.8*10 modes reacts with smeer Moles of ($16:25*103) (455410-5) (545-81103) 8 Feat = 31.25*10*meles =27.5*10**meles = 29x105Pmdes :./Meles of Fe?" Maus of Fe (31.25 *10*355-85) (27.5x10*255-85) (29x10^9255-85) sx Moles KMnO) que = 1746.9 1536 g. 1620 Atomie Maus of POD Fe = 55-85 ģ. (11 46x05 =1.6