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Magnetite can be reduced with $\mathrm{CO}$ to yield iron metal and carbon dioxide. Calculate the mass of magnetite (in $\mathrm{kg}$ ) needed to obtain $4 \mathrm{~kg}$ of iron if the process is $80 \%$ efficient. [Atomic weight of $\mathrm{Fe}$ and $\mathrm{O}$ are $56 \mathrm{~g}$ and $16 \mathrm{~g}$, respectively]
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The correct answer is:
6.9
$\underset{\text { Magnetite }}{\mathrm{Fe}_3 \mathrm{O}_4}+4 \mathrm{CO} \rightarrow \underset{4 \mathrm{~kg} \text { needed. }}{3 \mathrm{Fe}}+4 \mathrm{CO}_2$
$\mathrm{MW}$ of $\mathrm{Fe}_3 \mathrm{O}_4=232 \mathrm{~g} / \mathrm{mol}$
since the process has $80 \%$ efficiency then
$232 \mathrm{~g}$ of magnetite is producing $3 \times 56 \times 0.8 \mathrm{~g}$ of Fe
$\therefore \quad$ To obtain $4 \mathrm{~kg}$ of $\mathrm{Fe}$, the amount of $\mathrm{Fe}_3 \mathrm{O}_4$ required
$$
=\frac{232 \times 4 \times 10^3}{3 \times 56 \times 8} \mathrm{~g}=6.9 \mathrm{~kg}
$$
$\mathrm{MW}$ of $\mathrm{Fe}_3 \mathrm{O}_4=232 \mathrm{~g} / \mathrm{mol}$
since the process has $80 \%$ efficiency then
$232 \mathrm{~g}$ of magnetite is producing $3 \times 56 \times 0.8 \mathrm{~g}$ of Fe
$\therefore \quad$ To obtain $4 \mathrm{~kg}$ of $\mathrm{Fe}$, the amount of $\mathrm{Fe}_3 \mathrm{O}_4$ required
$$
=\frac{232 \times 4 \times 10^3}{3 \times 56 \times 8} \mathrm{~g}=6.9 \mathrm{~kg}
$$
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