Iron by Titration - Redox Rxn
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mlmusagi
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Iron by Titration - Redox Rxn


I am having trouble wrapping my head around this problem, how do I even get started?
5 Fe+2 +MnO4 + 8H+ = 5Fe+3 +Mn+2 + 4H2O
Suppose 1.507g of FeCl2 is placed in a 250 ml Erlenmeyer flask to which 50 mL of water and 10 mL of H2SO4 are added. This solution will be titrated by adding 0.1227 M KMnO4 . I would like to figure out moles of Fe+2 are initially present.
Do you use a new equation with FeCl2 + KMnO4 + H2O + H2SO4? Need something to make sense so I can work though this. Couldn't figure out the superscripts, sorry!
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ChenBeier
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1 mol FeCl2 corresponds to 1 mol Fe2+

n = m/M n = 1,507 g/126,75 g/mol = 0,0119 mol

This the molarity of Fe2+

0,12227 M KMnO4 = 0,12227 mol/l MnO4-

Calculate with this.
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mlmusagi
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thanks, the whole Cl2 addition was confusing me.
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