Mass of KHC2O4 (potassium acid oxalate) required to reduce 100 ml of 0.02 M KMnO4 in acidic medium (to Mn2+) is x g and to neutralise 100 ml of 0.05 M Ca(OH)2 is y g, then

Question

Mass of KHC2O4 (potassium acid oxalate) required to reduce 100 ml of 0.02 M KMnO4 in acidic medium (to Mn2+) is x g and to neutralise 100 ml of 0.05 M Ca(OH)2 is y g, then (A) x=y (B) 2x=y (C) x = 2y (D) none is correct

Tardigrade Admin · Accepted Answer

The reaction is as follows: 5 HC 2 O 4-+2 MnO 4- arrow 2 Mn 2++10 CO 2 The number of moles of MnO 4-=0.1 × 0.02=0.002 mol This is equal to 0.005 mol of HC 2 O 4-ion. This corresponds to xg. The number of moles of Ca ( OH )2 is 0.1 × 0.05=0.005 mol They corresponds to 0.01 mol of hydrogen ions and 0.01 mol of HC 2 O 4 -. This is equal to g. Hence, ( x / y )=(0.005/0.01)=(1/2) or y =2 x Hence, the relation is 2 x = y.

Q. Mass of $KH C_{2} O_{4}$ (potassium acid oxalate) required to reduce 100 ml of 0.02 M $K M n O_{4}$ in acidic medium (to $M n^{2 +}$ ) is x g and to neutralise $100$ ml of $0.05$ M Ca $(O H)_{2}$ is y g, then

$x = y$

$2 x = y$

$x = 2 y$

none is correct

Q. Mass of KHC2​O4​ (potassium acid oxalate) required to reduce 100 ml of 0.02 M KMnO4​ in acidic medium (to Mn2+) is x g and to neutralise 100 ml of 0.05 M Ca(OH)2​ is y g, then

x=y

2x=y

x=2y

none is correct

Q. Mass of $KH C_{2} O_{4}$ (potassium acid oxalate) required to reduce 100 ml of 0.02 M $K M n O_{4}$ in acidic medium (to $M n^{2 +}$ ) is x g and to neutralise $100$ ml of $0.05$ M Ca $(O H)_{2}$ is y g, then

$x = y$

$2 x = y$

$x = 2 y$