Question

100 tons of $Fe_2O_3$ containing 20% impurities will give iron by reduction with $H_2$ equal to

• A. 112 tons
• B. 80 tons
• C. 160 tons
• D. 56 tons

Answer 1

Answer: (D)

56 tons

Answer 2

$\underset{\text{160 g}}{ {Fe_{2}O_{3}}}+3H_{2} \to \underset{\text{112 g}}{ {2Fe}}+3H_{2}$ $Fe_{2}O_{3}$ given $=100$ tons $-20$ tons impurities $=80$ tons Fe obtained $=\frac{112}{160}\times80=56$ tons