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Grade: 12th pass
        
60 ml 0.1 M KMnO4 is treated with excess FeC2O4 in presence of H2SO4.The volume of CO2 gas (measured at NTP) obtained is :
a.  448 ml
b.  672 ml
c.  224 ml
d.  112 ml
please give answer in detail.....
8 months ago

Answers : (1)

ashutosh
15 Points
							
3KMnO4 + 5Fe2C2O4 + 24H+ ==> 5Fe3+ + 10 CO2 + 3Mn2+ + 3K+ + 12H2O.
This is the complete reaction in which KMnO4 oxidizes Fe2+. So now using mole concept we can sole this problem.
Moles of KMnO4 = 60*0.1/1000 = 6/1000
since 3 moles of Kmno4 gives 10 moles of CO2, thus 6/1000 moles of KMnO4 will give = 10*6/3*1000 = 2/100 moles
now, at NTP amount of 1 mol of CO2 = 22400 ml
So amount of 2/100 mol of CO2  = 2*22400/100 = 448 ml
so answer is (a)
 
8 months ago
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