a KMnO₄ + b H₂SO₄ + c FeSO₄ → K₂SO₄ + MnSO₄ + Fe₂(SO₄)₃+ H₂O. In this unbalanced stoichiometric equation the values of a,b and c respectively are

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2, 8, 10

1, 4, 10

2, 10, 8

2, 8, 16

answer is A.

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Detailed Solution

a KMnO₄ + bH₂SO₄ + cFeSO₄ → K₂SO₄ + MnSO₄ + Fe₂(SO₄)₃ + 3H₂O converting this equation to ionic equation (K⁺,SO₄^-2 are spectator ions) spectator ions due to undergo met chemical change during the process
$large MnO_4^ - ; + ;F{e^{ + 2}};xrightarrow{{;;{H^ + };;}};M{n^{ + 2}}; + ;F{e^{ + 3}}$
$large left. {begin{array}{*{20}{c}} {RHR} \ {MnO_4^ - ; to ;M{n^{ + 2}}} \ {Balancing;'O';atom} \ {MnO_4^ - ; to ;M{n^{ + 2}}; + ;4{H_2}O} \ {Balancing;'H';atom;in;acidic;medium} \ {MnO_4^ - ; + ;8{H^ + }; to ;M{n^{ + 2}}; + ;4{H_2}O} \ {Balancing;charg es} \ {MnO_4^ - ; + ;8{H^ + }; + ;5bar e; to ;M{n^{ + 2}}; + ;4{H_2}O} end{array}} right|begin{array}{*{20}{c}} {OHR} \ {F{e^{ + 2}}; to ;F{e^{ + 3}}} \ {Balancing;charg es} \ {F{e^{ + 2}}; to ;F{e^{ + 3}}; + ;bar e} \ end{array}$
$large begin{array}{*{20}{c}} {(OHR) times 5;;;;;;;;;;;;;;;;;;;;;;5F{e^{ + 2}}; to ;5F{e^{ + 3}}; + ;5bar e;;;;;;;;;;;;;;;;;;;;;;;} \ {(RHR);;;;;;;;;;;;;;;;;;;MnO_4^ - ; + ;8{H^ + }; + ;5bar e; to ;M{n^{ + 2}}; + ;4{H_2}O;;;;;;;} \ {overline {underline {Net;reaction;MnO_4^ - ; + ;5F{e^{ + 2}}; + ;8{H^ + }; to ;5F{e^{ + 3}}; + ;M{n^{ + 2}}; + ;4{H_2}O} } } end{array}$
mole ratio MnO₄^- :Fe⁺² 1:5
from the options 2:10
$large therefore$ a,b,c are 2,8 and 10