. Iron ore is impure, $\mathrm{Fe} 2 \mathrm{O} 3$. When $\mathrm{Fe} 2 \mathrm{O} 3$ is heated with an excess of carbon (coke), metallic iron and carbon monoxide gas are produced. From a sample of ore weighing $938 \mathrm{~kg}, 523 \mathrm{~kg}$ of pure iron is obtained. What is the mass percent $\mathrm{Fe} 2 \mathrm{O} 3$, by mass, in the ore sample, assuming that none of the impurities contain $\mathrm{Fe}$.

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∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation**: The reaction of iron(III) oxide with carbon to produce iron and carbon monoxide can be represented by the balanced chemical equation: $ \mathrm{Fe_2O_3} + 3 \mathrm{C} ightarrow 2 \mathrm{Fe} + 3 \mathrm{CO} $. ‖ ‖ 2 ⋮ **Calculate the molar mass of $\mathrm{Fe_2O_3}$ and $\mathrm{Fe}$**: The molar mass of $\mathrm{Fe_2O_3}$ is $(2 imes 55.85) + (3 imes 16.00) = 159.7 \, \mathrm{g/mol}$. The molar mass of $\mathrm{Fe}$ is $55.85 \, \mathrm{g/mol}$. ‖ ‖ 3 ⋮ **Determine the moles of $\mathrm{Fe}$ produced**: Convert the mass of pure iron obtained to moles using its molar mass: $523 \, \mathrm{kg} imes \frac{1000 \, \mathrm{g}}{1 \, \mathrm{kg}} imes \frac{1 \, \mathrm{mol}}{55.85 \, \mathrm{g}} = 9361.5 \, \mathrm{mol}$. ‖ ‖ 4 ⋮ **Use stoichiometry to find moles of $\mathrm{Fe_2O_3}$**: According to the balanced equation, 1 mole of $\mathrm{Fe_2O_3}$ produces 2 moles of $\mathrm{Fe}$. Therefore, the moles of $\mathrm{Fe_2O_3}$ used are $9361.5 \, \mathrm{mol} \div 2 = 4680.75 \, \mathrm{mol}$. ‖ ‖ 5 ⋮ **Calculate the mass of $\mathrm{Fe_2O_3}$ in the ore**: Convert the moles of $\mathrm{Fe_2O_3}$ back to mass: $4680.75 \, \mathrm{mol} imes 159.7 \, \mathrm{g/mol} = 747,119.25 \, \mathrm{g} = 747.119 \, \mathrm{kg}$. ‖ ‖ 6 ⋮ **Determine the mass percent of $\mathrm{Fe_2O_3}$**: The mass percent of $\mathrm{Fe_2O_3}$ in the ore is calculated by dividing the mass of $\mathrm{Fe_2O_3}$ by the total mass of the ore sample and multiplying by 100: $\frac{747.119 \, \mathrm{kg}}{938 \, \mathrm{kg}} imes 100 = 79.66\%$. ‖ ∻Answer∻ ⚹ The mass percent of $\mathrm{Fe_2O_3}$ in the ore sample is 79.66%. ⚹ ∻Key Concept∻ ⚹ Stoichiometry and molar mass are used to relate the mass of reactants to products in a chemical reaction. ⚹ ∻Explanation∻ ⚹ The mass percent of $\mathrm{Fe_2O_3}$ in the ore is found by using stoichiometry to calculate the mass of $\mathrm{Fe_2O_3}$ that would produce the given mass of iron, and then comparing this mass to the total mass of the ore sample◊What is the chemical equation for the reaction between $\mathrm{Fe} 2 \mathrm{O} 3$ and carbon to produce metallic iron and carbon monoxide gas?⍭ Generate me a similar question◊⚹

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