Question

It is desired to produce 2,000 kg of Fe in the reduction of its ore Fe₂O₃ using C as the reducing agent. What is the amount of C needed? How many Kg of the ore must be refined or reduced?

The equation: 2Fe₂O_3   +  3C = → 4Fe ₊ 3CO₂ 

Answer 1

M(Fe) = 55.84 g/mol

n(Fe) = $\frac{2000 \times 10^3}{55.84} = 35.51 \times 10^3 \;mol$

n(Fe₂O₃) $= \frac{1}{2}n(Fe)$

$= \frac{1}{2} \times 35.51 \times 10^3 \\ = 17.91 \times 10^3 \;mol$

M(Fe₂O₃) = 159.69 g/mol

m(Fe₂O₃) $= 17.91 \times 10^3 \times 159.69$

$= 2859.77 \times 10^3 \;g \\ = 2858.77 \;kg$

$n(C) = \frac{3}{4}n(Fe) \\ = \frac{3}{4} \times 35.51 \times 10^3 \\ = 26.63 \times 10^3 \;mol \\ m(C) = 26.63 \times 10^3 \times 12.01 \\ = 316.82 \times 10^3 \;g \\ = 316.82 \;kg$