Two tones of iron ore are containing 94% of Fe2O3 produces iron in pure state . Calculate the weight of iron.

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Solution

The atomic weight of $F e_{2} O_{3} = 159.69 g$
Atomic weight of 2 Fe atoms $= 2 \times 55.845 = 111.69$
The two tonnes of 94% pure $F e_{2} O_{3} = 2 \times 0.84 t o n n e s o f F e_{2} O_{3} = 1.88 t o n n e s o f 100 % p u r e F e_{2} O_{3}$
Hence, the weight of iron
$= \frac{M a s s o f F e i n F e_{2} O_{3}}{M a s s o f F e_{2} O_{3}}$
$= \frac{(2 \times 55.845)}{159.69} \times 1.88 = 1.315 t o n n e s = 1315 k g$
Hence, the weight of iron in two tonnes of iron ore with 94% of $F e_{2} O_{3}$ is 1315 kg

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