`Al_2 O_3 ` is reduced by electrolysis at ow potentials and high currents, If `4.0 xx 10^(4)` amperes of current is passed through molten `Al_2O_3` fro `6` hours , what mass of alumininum is produced ? (Assume ` 100%` current efficiency gt At mass of ` Al = 27 g ` "mol"^(-) )`.

Options

A. `1.3 xx 10^4 G`
B. `9.0 xx 10^3 g`
C. `8. 05 xx 10^4 g`
D. `2.4xx10^(5) g`

Answer

C

Explanation

Eq. of Al formed `= M/F = (it)/( 96500)`
` :. M_(Al) = (4xx 10^4 xx 6 xx 60xx 60xx 27)/( 96500 xx 3) (E = (27)/3)`
`:. m_(Al) = 8.05 xx 10^4 g`.