Iron by Titration - Redox Rxn

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mlmusagi
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Iron by Titration - Redox Rxn

Post by mlmusagi »

I am having trouble wrapping my head around this problem, how do I even get started?

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5 Fe+2 +MnO4 + 8H+ = 5Fe+3 +Mn+2 + 4H2O
Suppose 1.507g of

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FeCl2
is placed in a 250 ml Erlenmeyer flask to which 50 mL of water and 10 mL of

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H2SO4
are added. This solution will be titrated by adding 0.1227 M

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KMnO4
. I would like to figure out moles of

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Fe+2
are initially present.
Do you use a new equation with

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 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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Location: Berlin, Germany

Post by ChenBeier »

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.
mlmusagi
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Posts: 2
Joined: Sun Oct 22, 2017 11:31 am

Post by mlmusagi »

thanks, the whole Cl2 addition was confusing me.
phongphanp
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Joined: Tue Jan 02, 2018 1:02 am

Post by phongphanp »

I'm think this website focus on the small process . Limit on minimun step on calculation or specific on well-known chemistry equation .
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