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Chapter 3: Problem 99

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What masses of iron(III) oxide and aluminum must be used to produce \(15.0 \mathrm{~g}\) iron? What is the maximum mass of aluminum oxide that could be produced?

Short Answer

Expert verified
To produce 15.0 g of iron, 13.6 g of iron(III) oxide and 16.20 g of aluminum must be used. The maximum mass of aluminum oxide that could be produced is 13.7 g.

Step by step solution

01

Convert mass of iron to moles

Since the desired amount of iron is 15.0 g, we first need to convert that mass to moles using the molar mass of iron, which is approximately 55.85 g/mol. Moles of iron = \(\frac{15.0 \ \text{g}}{55.85 \ \text{g/mol}}\)
02

Determine moles of reactants and products using stoichiometry

From the balanced chemical equation, we can see the stoichiometry of the reaction: \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2\mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) For every 2 moles of iron produced, 1 mole of iron(III) oxide and 2 moles of aluminum are required (also 1 mole of aluminum oxide is formed). We can now calculate the moles of iron(III) oxide, aluminum, and aluminum oxide using the stoichiometry and the moles of iron obtained in step 1. Moles of iron(III) oxide = \(\frac{1}{2}\) moles of iron Moles of aluminum = moles of iron Moles of aluminum oxide = \(\frac{1}{2}\) moles of iron
03

Convert moles of reactants and products to mass

Next, we need to convert the moles of iron(III) oxide, aluminum, and aluminum oxide to mass using their respective molar masses. The molar masses are: - Iron(III) oxide (Fe2O3): 159.69 g/mol - Aluminum (Al): 26.98 g/mol - Aluminum oxide (Al2O3): 101.96 g/mol Mass of iron(III) oxide = moles of iron(III) oxide × molar mass of iron(III) oxide Mass of aluminum = moles of aluminum × molar mass of aluminum Mass of aluminum oxide = moles of aluminum oxide × molar mass of aluminum oxide After calculating the masses using the molar masses, we get the required masses of iron(III) oxide and aluminum to produce 15.0 g of iron, and the maximum mass of aluminum oxide that could be produced.

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Most popular questions from this chapter

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains \(91.27 \% \mathrm{E}\) and \(8.73 \% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{\mathrm{s}}\), calculate the atomic mass of \(\mathrm{E}\).

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 \(0.6498 \mathrm{~g}\) of one of the compounds leaves a residue of \(0.6018 \mathrm{~g}\). Heating \(0.4172 \mathrm{~g}\) of the other compound results in a mass loss of \(0.016 \mathrm{~g}\). Determine the empirical formula of each compound.

Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH \((188.35 \mathrm{~g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{~g} / \mathrm{mol})\) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{~g} / \mathrm{mol})\) d. SN \((184.32 \mathrm{~g} / \mathrm{mol})\)

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g}\) \(\mathrm{CO}\), and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{~g} \mathrm{~N}_{2}\) and \(5.00 \times 10^{2} \mathrm{~g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?
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