Explanation

(d)
In Hall and Heroult process
$\ \ \ \ \ \ \ \ \ \ 2Al_{2}O_{3} \longrightarrow 4Al + 3O_{2}$
$\ \ \ \ \ \ 4C + 3O_{2} \longrightarrow 2CO_{2} + 2CO \uparrow$
$2Al_{2}O_{3} + 4C \longrightarrow 4Al + 2CO_{2} + 2CO$
Only for removal of $CO_{2}$, following equation is possible
$2Al_{2}O_{3} + 3C \longrightarrow 4Al + 3CO_{2}$
$3 \times 12$ $4 \times 27$
=36 g =108 g
$\because$ For 108 g of Al, 36 g of C is required in above reaction.
$\therefore$ For $270 \times 10^{3}$g of Al, required amount of C
$= \frac{36}{108} \times 270 \times 10^{3}$
$= 90\ kg$