Explanation

(d)
In 100 tons of $Fe_{2}O_{3},$ pure $Fe_{2}O_{3}$
$= 100 - \frac{100 \times 20}{100}$
$= 80$ tons
$Fe_{2}O_{3} + 3H_{2} \longrightarrow 2Fe_{2} + 3H_{2}O$
$2 \times 56 + 48$ $2 \times 56$
160 $2 \times 56$
$\because$ 160 g $Fe_{2}O_{3}$ gives $Fe = 2 \times 56$ g
$\therefore$ 80 tons $Fe_{2}O_{3}$ will give $Fe = \frac{2 \times 56 \times 80}{160}$
= $56$ tons