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Chapter 3: Problem 157

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 -g sample of impure iron ore is heated with excess carbon, producing 453 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 86.08%.

Step by step solution

01

Calculate the moles of pure iron produced

In order to determine the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the sample, we need to first find the moles of pure iron produced by the reaction. To do that, we will use the molar mass of iron which is 55.85 g/mol $$ \text{moles of pure iron} = \frac{\text{mass of pure iron}}{\text{molar mass of iron}} = \frac{453 \,\text{g}}{55.85 \,\text{g/mol}} = 8.11 \,\text{mol} $$
02

Calculate the moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the sample

From the balanced chemical equation, we know that 1 mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) produces 2 moles of iron. Using stoichiometry, we can find the moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the sample: $$ \text{moles of} \,\mathrm{Fe}_{2} \mathrm{O}_{3} = \frac{\text{moles of pure iron}}{2} = \frac{8.11 \,\text{mol}}{2} = 4.055 \,\text{mol} $$
03

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

Now that we have the moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the sample, we can find its mass by using the molar mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) which is 159.69 g/mol: $$ \text{mass of}\,\mathrm{Fe}_{2} \mathrm{O}_{3} = \text{moles of}\,\mathrm{Fe}_{2} \mathrm{O}_{3} \times \text{molar mass of}\,\mathrm{Fe}_{2} \mathrm{O}_{3} = 4.055 \,\text{mol} \times 159.69\,\text{g/mol} = 647.47 \,\text{g} $$
04

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

Now that we have the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the sample, we can calculate its mass percent by dividing it by the total mass of the impure iron ore sample and multiplying the result by 100% $$ \text{mass percent of}\, \mathrm{Fe}_{2} \mathrm{O}_{3} = \frac{\text{mass of}\, \mathrm{Fe}_{2} \mathrm{O}_{3}}{\text{mass of impure iron ore sample}} \times 100\% = \frac{647.47 \, \text{g}}{752 \,\text{g}} \times 100\% = 86.08\% $$ The mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample is 86.08%.

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Most popular questions from this chapter

Balance each of the following chemical equations. a. $\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)$ b. $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ c. $\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ d. $\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)$ e. $\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)$ f. $\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MoO}_{3}(s)+\mathrm{SO}_{2}(g)$ g. $\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)$

Fungal laccase, a blue protein found in wood-rotting fungi, is 0.390$\% \mathrm{Cu}$ by mass. If a fungal laccase molecule contains four copper atoms, what is the molar mass of fungal laccase?

Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed as NutraSweet. The molecular formula of aspartame is $\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{N}_{2} \mathrm{O}_{5}$ a. Calculate the molar mass of aspartame. b. What amount (moles) of molecules are present in 10.0 \(\mathrm{g}\) aspartame? c. Calculate the mass in grams of 1.56 mole of aspartame. d. What number of molecules are in 5.0 mg aspartame? e. What number of atoms of nitrogen are in 1.2 g aspartame? f. What is the mass in grams of \(1.0 \times 10^{9}\) molecules of aspartame? g. What is the mass in grams of one molecule of aspartame?

The most common form of nylon (nylon-6) is 63.68\(\%\) carbon, 12.38\(\%\) nitrogen, 9.80\(\%\) hydrogen, and 14.14\(\%\) oxygen. Calculate the empirical formula for nylon-6.

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating 0.6498 g of one of the compounds leaves a residue of 0.6018 g. Heating 0.4172 g of the other compound results in a mass loss of 0.016 g. Determine the empirical formula of each compound.
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