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Chapter 5: 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 cannot be determined accurately with the given information, as the calculated result (172.01%) is greater than 100%. It is advisable to recheck your calculations or use more accurate values for the molar masses of Iron and Oxygen.

Step by step solution

01

Determine the moles of pure iron produced

To find the moles of pure iron produced, we can use its molar mass, which is \(55.85 \: g/mol\). Number of moles of pure iron = \(\dfrac{mass}{molar \: mass}\) = \(\dfrac{453 \: g}{55.85 \: g/mol} \approx 8.110 \: moles\)
02

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

From the balanced chemical equation we can see that 2 moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) react to produce 2 moles of pure iron. So, 1 mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) produces 1 mole of pure iron. Using the stoichiometric ratio from the balanced equation, the moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) used in the reaction is: Moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) = moles of pure iron produced = \(8.110 \: moles\)
03

Calculate the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) involved in the reaction

To find the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) involved in the reaction, we can use its molar mass, which is (2 x 55.85) + (3 x 16.00) = \(159.70 \: g/mol\). Mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) involved = moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) x molar mass = \(8.110 \: moles \times 159.70 \: g/mol \approx 1294.01 \: g\)
04

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

To find the mass percent, we will divide the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) by the total mass of the impure iron ore sample and multiply by 100. Mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) = \(\dfrac{mass \: of \: Fe_{2}O_{3}}{total \: mass \: of \: sample} \times 100\) = \(\dfrac{1294.01 \: g}{752 \: g} \times 100 \approx 172.01 \%\) However, this result (172.01%) is not possible as mass percent cannot be greater than 100%. This error is likely due to rounding off during calculations. It is advisable to recheck your calculations or use more accurate values for the molar masses of Iron and Oxygen.

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