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Chapter 4: Problem 111

A 2.05 g sample of an iron-aluminum alloy (ferroaluminum) is dissolved in excess HCl(aq) to produce \(0.105 \mathrm{g} \mathrm{H}_{2}(\mathrm{g}) .\) What is the percent composition, by mass, of the ferroaluminum? $$\begin{array}{c} \mathrm{Fe}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{FeCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) \\ 2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g}) \end{array}$$

Short Answer

Expert verified
The percent composition by mass of the ferroaluminum is approximately 56.7% Fe and 41.2% Al.

Step by step solution

01

Calculate the Moles of Hydrogen

Using the mass of hydrogen produced and its molar mass, calculate the moles of hydrogen. Molar mass of hydrogen (H2) is 2.016 g/mol. Moles = Mass / Molar Mass => Moles = 0.105 g / 2.016 g/mol = 0.0521 mol.
02

Calculate Moles of Fe and Al

Use the stoichiometry of the reactions to calculate moles of Fe and Al. From the reactions, 1 mol of Fe produces 1 mol of H2, and 1 mol of Al produces 1.5 mol of H2. Divide the moles of H2 by 2.5 to calculate moles of Fe, and subtract this from total moles of H2 to find the moles of Al. Moles of Fe = 0.0521 mol / 2.5 = 0.0208 mol, Moles of Al = 0.0521 mol - 0.0208 mol = 0.0313 mol.
03

Calculate mass of Fe and Al

Use the molar masses of Fe and Al to determine their masses in the sample. Molar mass of Fe is 55.845 g/mol and molar mass of Al is 26.982 g/mol. Mass Fe = 0.0208 mol * 55.845 g/mol = 1.162 g, Mass Al = 0.0313 mol * 26.982 g/mol = 0.844 g.
04

Calculate Percent Composition

The percent composition is given by the individual masses of the elements divided by the total mass of the sample, all times 100%. Percent Fe = (1.162 g / 2.05 g) * 100% = 56.7%, Percent Al = (0.844 g / 2.05 g) * 100% = 41.2%.

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