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500mg of iron(ll) complex ferrous bisglycinate hydrochloride was dissolved in dilute H2SO4 and titrated with 0.0200 mol. dm3 KMnO4 18.10cm3 of KMnO4 solution were require to reach the end point the equation for the titration reaction is as follows 5Fe2+ + MnO4- + 8H+ - - > 5Fe2+ + Mn2+ +4H2O

Calculate the
1.number of moles of Fe2+ in the capsule
2.mass of iron in the capsul
3.molar mass of the iron (ll) complex assuming 1mole of the complex contains 1mole of iron. (Fe=55.9gmol-1) ​

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pstnonsonjoku

Using the stoichiometry of the reaction, we can obtain the molar mass of the complex as 276 g/mol. The number of moles iron II as 0.00181 moles and the mass of iron II as  0.1 g.

Redox reaction

A redox reaction is one in which there is loss or gain of electrons. The equation of the reaction is 5Fe2+ + MnO4- + 8H+ - - > 5Fe2+ + Mn2+ +4H2O

Number of moles of permanganate =   0.0200 mol. dm3 * 18.10cm3/1000 L

= 0.000362 moles

If 5 moles of iron II reacted with 1 mole of permanganate

x moles of iron II reacts with 0.000362 moles of permanganate

x = 5 moles * 0.000362 moles/ 1 mole

= 0.00181 moles of iron II

Recall that;

Number of moles = mass/molar mass

Molar mass of iron  II = 56 g/mol

0.00181 moles  = x/56 g/mol

x = 0.00181 moles * 56 g/mol = 0.1 g

If 1 mole of the complex contains 1 mole of iron then;

0.00181 moles = 500 * 10^-3/MM

MM= 500 * 10^-3/0.00181 moles

MM = 276 g/mol

Learn more about stoichiometry: https://brainly.com/question/9743981