Redox and Equivalent ConceptHard
Question
A certain amount of a reducing agent reduces x mole of MnO2 and y mole of K2CrO4 in different reactions in acidic medium. If the changes in oxidation states of reducing agent in the reactions are in 1: 2 ratio, respectively, then the ratio of x and y is
Options
A.2:3
B.1:3
C.3:4
D.3:2
Solution
$n_{eq}R.A. = n_{eq}MnO_{2} \Rightarrow n \times 1 = x \times 2\text{and }n_{eq}R.A. = n_{eq}K_{2}CrO_{4} \Rightarrow n \times 2 = y \times 3$
$\therefore x:y = 3:4$
Create a free account to view solution
View Solution FreeMore Redox and Equivalent Concept Questions
The volume of 0.10 M-AgNO3 should be added to 10.0 ml of 0.09 M-K2CrO4 to precipitate all the chromate as Ag2CrO4 is...Equivalent weight of a metal is 18.67. When it reacts with chlorine, the mass of metal which will form 162.52 g of metal...What volume of 0.40 M-Na2S2O3 would be required to react with the I2 liberated by adding excess of KI to 50 ml of 0.20 M...The formula of brown ring complex is [Fe(H2O)5(NO)]SO4. The oxidation state of iron is...Which of the following have been arranged in the order of decreasing oxidation number of sulphur?...