Question

A sample of feso4 and fec2o4 dissolved in h2so4. the complete oxidation of sample required 8./3 eq of kmno4. after oxidation, the reaction mixture was reduced by zn on again oxidation by kmmo4 required 5/3eq the mole ratio of feso4 and fec2o4?

Answer 1

Answer:

let equivalent of fec2o4=x

Explanation:

equivalent of kmno4= equivalent of fe+2 + equivalent of c2o4-2 + equivalent of fe+2

8/3=2x+equivalent of fe+2

equivalent of kmno4 = equivalent of fe+3  formed

5/3= x+ equivalent of fe+3

let equivalent of feso4=y

then equivalent of fe+3 formed = y

so

8/3 = 2x+y

5/3=x+y

x=1

and y=2/3

equivalent of feso4 = 2/3 and equivalent of fec2o4=1

moles*f=2/3

moles=1/3

moles*f=1

moles=1/2

moles ratio = 2/3

Answer 2

Given:

Sample of FeSO₄ and FeC₂O₄ dissolved in H₂SO₄ Then complete oxidation occurs followed by reduction. Again oxidation done by KMnO₄.

To find:

Mole ratio of FeSO₄ and FeC₂O₄.

Solution:

Lets assume the milli moles of FeSO₄ = a

and milli moles of FeC₂O₄ = b

Equivalent of KMnO₄ = 8/3 (given)

so, equivalents of FeC₂O₄ + equivalent of FeSO₄ =equivalent of KMnO₄

putting all the values equation formed is

 ${{\frac {{8}}{{3}}} = {{\text{{3}}{ \cdot a}}} + \left {{{b }}{{}}} \right$  …..(1)

now, due to reduction by zinc and H₂SO₄, only Fe⁺³ ions reduced to Fe⁺² again

Fe⁺² ions formed again oxidised by KMnO₄ .Therefore,

${{\frac {{5}}{{3}}} = {{\text{{a}}{ }}} + \left {{{b }}{{}}} \right$  ….(2)

by solving equation(1) and (2), we get values of a and b,

${{\text {{a}}} = {{\frac{{7}}{{6}}}$  and

${{\text {{b}}} = {{\frac{{1}}{{2}}}$

so, the mole ratio of FeSO₄ and FeC₂O₄ is 7:3.