20 " mL of " `(M)/(60)` `KBrO_3` was reacted with a sample of `SeO_(3)^(2-)` 20 The `Br_(2)` thus evolved was removed and the excess of `NaAsO_(2)` The reaction involved are
`SeO_(3)^(2-)+BrO_(3)^(ɵ)+H^(o+)toSeO_(4)^(2-)+Br_(2)+H_(2)O` .. (i)
`BrO_(3)^(ɵ)+ASO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+AsO_(4)^(3-)+H^(o+)` ..(ii)
`[Mw(SeO_(3)^(2-))=79+48=127gmol^(-1)]`
Q. Excess of m" Eq of "`BrO_(3)^(ɵ)` in reaction (ii) is

Options

A. `(1)/(6)`
B. `(11)/(6)`
C. `(1)/(36)`
D. `(11)/(36)`

Answer

A

Explanation

Total mmoles `BrO_(3)^(ɵ)=20xx(1)/(60)=(1)/(3)`
Total mmoles of `AsO_(2)^(ɵ)=(1)/(60)xx5=(1)/(12)`
According to the balanced equation for first and second reaction.
`5SeO_(3)^(2-)+2BrO_(3)^(ɵ)+2H^(o+)to5SeO_(4)^(2-)+Br_(2)+H_(2)O` .. (i)
`BrO_(3)^(ɵ)+3AsO_(2)^(ɵ)+H_(2)OtoBr^(ɵ)+3AsO_(4)^(3-)+6H^(o+)` .(ii)
`therefore 3 m" mol of "AsO_(2)^(ɵ)=1 m" mol of "BrO_(3)^(ɵ)` in equation (ii)
`(1)/(12)` m" mol of "`AsO_(2)^(ɵ)=(1)/(3)xx(1)/(12)`
`=(1)/(36)` m" mol of "excess `BrO_(3)^(ɵ)`
mmoles of `BrO_(3)^(ɵ)` reacted with `SeO_(3)^(2-)` in equation (i)
`=(1)/(3)-(1)/(36)=(11)/(36)` m" mol of "`BrO_(3)^(ɵ)` reacted
`[BrO_(3)^(ɵ)-=AsO_(2)^(ɵ)]` equation (ii)
`mEq-=-mEq`
`(1)/(36)mmol=(1)/(12)mmol`
`(1)/(36)xx6mEq=(1)/(12)xx2mEq`
`(1)/(6)` mEq excess m" Eq of "`BrO_(3)^(ɵ)-=(1)/(6)mEq`
Hence, answer is `(a)=(1)/(6)mEq`