Question

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. \(\mathrm{A}\) \(752-\mathrm{g}\) sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

Short answer

The mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample is approximately 86.10%.

Step-by-step solution

Calculate moles of pure iron produced

Given that the mass of pure iron produced is 453 g, we can find the moles of iron by using the molar mass of iron which is 55.85 g/mol. $$ \text{moles of Fe} = \frac{453 \, \text{g}}{55.85 \, \text{g/mol}} \approx 8.11 \, \text{moles} $$

Find moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) needed for the reaction

From the balanced chemical equation, we can see that 1 mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) produces 2 moles of iron: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s) + 3 \mathrm{CO}(g) $$ Therefore, we can find the moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) needed for the reaction: $$ \text{moles of } \mathrm{Fe}_{2} \mathrm{O}_{3} = \frac{1}{2} \times \text{moles of Fe} \approx \frac{1}{2} \times 8.11 \, \text{moles} \approx 4.055 \, \text{moles} $$

Convert moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) to mass in grams

Now we can find the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) needed for the reaction using its molar mass (159.69 g/mol): $$ \text{mass of } \mathrm{Fe}_{2} \mathrm{O}_{3} = 4.055 \, \text{moles} \times 159.69 \, \text{g/mol} \approx 647.3 \, \text{g} $$

Calculate mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample

Finally, we can find the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample: $$ \text{mass percent of } \mathrm{Fe}_{2} \mathrm{O}_{3} = \frac{\text{mass of } \mathrm{Fe}_{2} \mathrm{O}_{3}}{\text{mass of impure iron ore sample}} \times 100\% $$ $$ \text{mass percent of } \mathrm{Fe}_{2} \mathrm{O}_{3} = \frac{647.3 \, \text{g}}{752 \, \text{g}} \times 100\% \approx 86.10\% $$ Thus, the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample is approximately 86.10%.

Key concepts

Molar Mass

Understanding molar mass is key when dealing with chemical reactions. It represents the mass of one mole of a substance. For

• Elements, it's the atomic mass expressed in grams per mole (g/mol).
• Compounds, such as \( \text{Fe}_2\text{O}_3 \), it’s the sum of the atomic masses of its components.

To compute the molar mass of \( \text{Fe}_2\text{O}_3 \), add the atomic masses of two iron (Fe) atoms and three oxygen (O) atoms:- Molar mass of Fe: \(55.85 \, \text{g/mol} \)- Molar mass of O: \(16.00 \, \text{g/mol} \)Thus, molar mass of \( \text{Fe}_2\text{O}_3 \) is:\[2 \times 55.85 + 3 \times 16.00 = 159.69 \, \text{g/mol}\]This calculation helps in converting between grams and moles, a crucial step in reaction stoichiometry. Whenever you see a chemical equation, think about how molar mass allows us to relate masses to moles, and therefore to actual chemical quantities in reactions.

Pure Iron Production

Producing pure iron involves reducing iron oxide (\( \text{Fe}_2\text{O}_3 \)) using carbon. In this chemical process, carbon reacts with iron oxide to produce pure iron and carbon monoxide (\( \text{CO} \)). The balanced reaction is:\[\text{Fe}_2 \text{O}_3 (s) + 3 \text{C} (s) \rightarrow 2 \text{Fe} (s) + 3 \text{CO} (g)\]Here's how it works:- For every mole of \( \text{Fe}_2\text{O}_3 \) that reacts, two moles of \( \text{Fe} \) are produced.Understanding this reaction ratio is crucial. It tells us how much \( \text{Fe}_2\text{O}_3 \) is needed to produce a certain quantity of iron. In our case:- 453 g of iron corresponds to \( \approx 8.11 \text{ moles} \) of iron as calculated by dividing iron's mass by its molar mass \( (453 \, \text{g} / 55.85 \, \text{g/mol}) \).- With the reaction stoichiometry, this means we need about half, or \( 4.055 \text{ moles} \) of \( \text{Fe}_2\text{O}_3 \).

Mass Percent Calculation

Calculating mass percent helps us understand the composition of complex mixtures. We can find out how much of a specific component is present compared to the whole. Here’s how to do it:To determine the mass percent of a component in a mixture:1. Identify the mass of the component (in grams).2. Identify the total mass of the mixture (in grams).3. Divide the component's mass by the total mass and multiply by 100 to get the percentage.For our case study:- Mass of \( \text{Fe}_2\text{O}_3 \) needed is calculated to be \( 647.3 \, \text{g} \).- Total mass of the iron ore sample is \( 752 \, \text{g} \). The mass percent of \( \text{Fe}_2\text{O}_3 \) is:\[\frac{647.3}{752} \times 100 \% \approx 86.10\%\]This calculation shows a high purity level of iron oxide in the ore, informing us about the quality of the sample used for producing iron.