Explanation

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