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Newbie

Joined: 22 Oct 2017
Posts: 2

# 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!

Sr. Staff Member

Joined: 27 Sep 2017
Posts: 159
Location: Berlin, Germany
 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.

Newbie

Joined: 22 Oct 2017
Posts: 2
 thanks, the whole Cl2 addition was confusing me.

Sr. Member

Joined: 02 Jan 2018
Posts: 33
 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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