the volume in ml of 0.05 M KMnO4(acidic) required to oxidize 2 gm of FeSO4 in dilute solution is
is ans 52.63 ml ?
KMnO4 + 5FeSO4 ==== 5 FE3+ + Mn2+
oxidation no. of Mn changes frm +7 to +2 gain of 5e-
therefore 5 Fe reacts with KMnO4 so that Fe2+ changes to Fe3+
e- loss == e- gain
therefore 1 mol requires 5 mol of Fe2+
so ? many moles of KMno4 shud react with 2/152 mols of Fe (mol wt of FeSO4 is 152)
it cums out 2 b 2.63 milli moles
derfore 0.05 * vol = 2.63 mili mols
so vol is 52.6 ml
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