Magnetite can be reduced with CO to yield iro | vedclass

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(D) The balanced chemical equation for the reduction of magnetite $(Fe_3O_4)$ is: $Fe_3O_4 + 4CO \rightarrow 3Fe + 4CO_2$ Molar mass of $Fe_3O_4 = (3

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$6.9$

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(D) The balanced chemical equation for the reduction of magnetite $(Fe_3O_4)$ is: $Fe_3O_4 + 4CO ightarrow 3Fe + 4CO_2$ Molar mass of $Fe_3O_4 = (3 	imes 56) + (4 	imes 16) = 168 + 64 = 232 \ g/mol$. From the stoichiometry, $232 \ g$ of $Fe_3O_4$ produces $3 	imes 56 = 168 \ g$ of $Fe$ at $100 \%$ efficiency. Given the process is $80 \%$ efficient, the actual yield of $Fe$ from $232 \ g$ of $Fe_3O_4$ is $168 	imes 0.8 = 134.4 \ g$. To obtain $4 \ kg$ $(4000 \ g)$ of $Fe$, the mass of $Fe_3O_4$ required is: $	ext{Mass} = \frac{232 	imes 4000}{134.4} \approx 6904.76 \ g \approx 6.9 \ kg$.

Magnetite can be reduced with $CO$ to yield iron metal and carbon dioxide. Calculate the mass of magnetite (in $kg$) needed to obtain $4 \ kg$ of iron if the process is $80 \%$ efficient. [Atomic weight of $Fe$ and $O$ are $56 \ g$ and $16 \ g$, respectively]

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