`4.9 g` of `K_(2)Cr_(2)O_(7)` is taken to prepare `0.1 L` of the solutio. `10 mL` of this solution is further taken to oxidise `Sn^(2+)` ion into `Sn^(4+) ion` so produced is used in second reaction to prepare `Fe^(3+)` ion then the millimoles of `Fe^(3+)` ion formed will be (assume all other components are in sufficient amount)[Molar mass of `K_(2)Cr_(2)O_(7)=294 g`].

Options

A. 5
B. 20
C. 10
D. 15

Answer

C

Explanation

N of `K_(2)Cr_(2)O_(7) = (4.9 xx 1)/(49xx 0.1) = 1N`
No. of m.equts orf `Fe^(+3) =` No. of m.eqts of `Sn^(4+)`
=No. of m.equts of `K_(2)Cr_(2)O_(7) = 10 xx 1 = 10`