4.55 g Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 mL of 56 vol strength of H2O2, then(prenominal) what will be the section purity of Ba(MnO4)2 in the sample ? (Ba 137, Mn 55)(A) 40%(B)25%(C) 10%(D) 68.18%
Note-
I think you have written it wrong the correct question is 55 g of Ba(MnO4)2 not 4.55 g
Formula used -
Volume strength of H2O2= 5.6 x Normality
56 vol strength of H2O2 = 5.6 x Normality
Normality of = 56/5.6 = 10
Reaction Involved-
Ba(MnO4)2 + H2O2 = Ba(OH)2 + 2 MnO2 + 2 O2
In terms of normality,
Milli Equivalents of Ba(MnO4)2 reacted = Equivalents of H2O2
= 10 x 100
=1000 mili equivalent
= 1 equivalent
As Milli Equivalents of Ba(MnO4)2 reacted = Equivalents of H2O2
Therefore,
Moles of Ba(MnO4)2 = 1/10 = 0.1
Weight of Ba(MnO4)2 = 0.1 x Molecular mass of Ba(MnO4)2
Weight of Ba(MnO4)2 = 0.1 x 375
% purity of Ba(MnO4)2 in the sample
= (0.1 x 375)/55
= (0.6818) x 100
= 68.18 %
Answer
I think you have written it wrong the correct question is 55 g of Ba(MnO4)2 not 4.55 g
Formula used -
Volume strength of H2O2= 5.6 x Normality
56 vol strength of H2O2 = 5.6 x Normality
Normality of = 56/5.6 = 10
Reaction Involved-
Ba(MnO4)2 + H2O2 = Ba(OH)2 + 2 MnO2 + 2 O2
In terms of normality,
Milli Equivalents of Ba(MnO4)2 reacted = Equivalents of H2O2
= 10 x 100
=1000 mili equivalent
= 1 equivalent
As Milli Equivalents of Ba(MnO4)2 reacted = Equivalents of H2O2
Therefore,
Moles of Ba(MnO4)2 = 1/10 = 0.1
Weight of Ba(MnO4)2 = 0.1 x Molecular mass of Ba(MnO4)2
Weight of Ba(MnO4)2 = 0.1 x 375
% purity of Ba(MnO4)2 in the sample
= (0.1 x 375)/55
= (0.6818) x 100
= 68.18 %
Answer