100 tons of Fe2O3 containing 20 % impurities will give iron by reduction with H2 equal to

Question

100 tons of Fe2O3 containing 20 % impurities will give iron by reduction with H2 equal to (A) 112 tons (B) 80 tons (C) 160 tons (D) 56 tons

Tardigrade Admin · Accepted Answer

underset160 g Fe 2 O 3+3 H 2 arrow underset112 g 2 Fe +3 H 2 Fe 2 O 3 given =100 tons -20 tons impurities =80 tons Fe obtained =(112/160) × 80=56 tons

Q. $100$ tons of $F e_{2} O_{3}$ containing $20%$ impurities will give iron by reduction with $H_{2}$ equal to

112 tons

80 tons

160 tons

56 tons

$160 g F e_{2} O_{3} + 3 H_{2} \to 112 g 2 F e + 3 H_{2}$
$F e_{2} O_{3}$ given $= 100$ tons $- 20$ tons impurities $= 80$ tons
Fe obtained $= \frac{112}{160} \times 80 = 56$ tons

Q. 100 tons of Fe2​O3​ containing 20% impurities will give iron by reduction with H2​ equal to