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Chapter 3: Problem 147

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

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

Step by step solution

01

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. $$ \text{moles of Fe} = \frac{453 \, \text{g}}{55.85 \, \text{g/mol}} \approx 8.11 \, \text{moles} $$
02

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: $$ \text{moles of } \mathrm{Fe}_{2} \mathrm{O}_{3} = \frac{1}{2} \times \text{moles of Fe} \approx \frac{1}{2} \times 8.11 \, \text{moles} \approx 4.055 \, \text{moles} $$
03

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): $$ \text{mass of } \mathrm{Fe}_{2} \mathrm{O}_{3} = 4.055 \, \text{moles} \times 159.69 \, \text{g/mol} \approx 647.3 \, \text{g} $$
04

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%.

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Most popular questions from this chapter

Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of \(47.6 \mathrm{mg}\) cumene produces some \(\mathrm{CO}_{2}\) and \(42.8 \mathrm{mg}\) water. The molar mass of cumene is between 115 and \(125 \mathrm{~g} / \mathrm{mol}\). Determine the empirical and molecular formulas.

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