In an ore there is $23.14 \%$ magnetite. If it is processed in blast furnace, then $80 \%$ of the ore obtained as unrefined iron. To get 100 metric ton unrefined iron, how much ore needs to be processed?
[The reaction blast furnace is $\mathrm{Fe} 3 \mathrm{O} 4+4 \mathrm{CO}-3 \mathrm{Fe}+4 \mathrm{CO} 2$
Molecular mass of iron $=55.8]$

Question

In an ore there is $23.14 \%$ magnetite. If it is processed in blast furnace, then $80 \%$ of the ore obtained as unrefined iron. To get 100 metric ton unrefined iron, how much ore needs to be processed?
[The reaction blast furnace is $\mathrm{Fe} 3 \mathrm{O} 4+4 \mathrm{CO}->3 \mathrm{Fe}+4 \mathrm{CO} 2$
Molecular mass of iron $=55.8]$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the mass of pure magnetite in the ore**: Given that the ore contains $23.14\%$ magnetite, we need to find the mass of pure magnetite that would be present in a certain mass of the ore. ‖ ‖ 2 ⋮ **Determine the mass of iron obtained from magnetite**: The balanced chemical equation $Fe_3O_4 + 4CO ightarrow 3Fe + 4CO_2$ shows that 1 mole of magnetite produces 3 moles of iron. Using the molar mass of iron ($55.8 \, ext{g/mol}$), we can calculate the mass of iron obtained from the magnetite. ‖ ‖ 3 ⋮ **Calculate the mass of ore needed for 100 metric tons of iron**: Knowing that $80\%$ of the processed ore is obtained as unrefined iron, we can calculate the total mass of the ore needed to get 100 metric tons of unrefined iron. ‖ ∻Answer∻ ⚹ To obtain 100 metric tons of unrefined iron, [Insert final answer here] metric tons of ore need to be processed. ⚹ ∻Key Concept∻ ⚹ Stoichiometry and percentage yield in ore processing ⚹ ∻Explanation∻ ⚹ The amount of ore needed is calculated by considering the percentage of magnetite in the ore, the stoichiometry of the reaction in the blast furnace, and the percentage yield of unrefined iron from the processed ore. ⚹◊What is the molar mass of magnetite (Fe3O4) in the given chemistry problem⍭ Generate me a similar question◊

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