calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide?

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Solution

$T h e a m o u n t o f i r o n p r e s e n t i n F e_{2} O_{3} w h o s e m a s s i s 160 g = 2 \times 56 = 112 g S o i n 10 K g (10 \times 10^{3} g) o f F e_{2} O_{3} = \frac{112}{160} \times 10 \times 10^{3} = 7 \times 10^{3} g F o r 100 % p u r i t y o f 10 K g t h e a m o u n t o f i r o n = 7 \times 10^{3} g F o r 80 % p u r i t y = \frac{7 \times 10^{3}}{100} \times 80 = 5.6 \times 10^{3} g$

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