Problem 111 - General Chemistry: Principles and Modern Applications

Question

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? $$\begin{array}{c} \mathrm{Fe}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{FeCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) \\ 2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g}) \end{array}$$

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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%.

Key Concepts

Stoichiometry

Stoichiometry is an essential tool in chemistry that helps us connect the amounts of reactants and products involved in a chemical reaction. It allows us to understand how quantities of different substances are related to each other through their chemical ratios.
For the given reactions involving iron (\( ext{Fe} \)) and aluminum (\( ext{Al} \)) with hydrochloric acid (\( ext{HCl} \)), stoichiometry helps us determine the proportion of hydrogen gas (\( ext{H}_2 \)) produced by each metal. In one reaction, 1 mole of iron produces 1 mole of hydrogen gas. In the reaction with aluminum, 1 mole of aluminum produces 1.5 moles of hydrogen gas.
Understanding these relationships is crucial for calculating how much of each metal contributes to the production of hydrogen gas, and ultimately helps determine the percent composition of the mixture.

Molar Mass Calculation

Calculating molar mass is a key step in many chemical calculations. Molar mass refers to the mass of one mole of a given substance and is typically measured in grams per mole (g/mol).
To calculate the molar mass of hydrogen gas (\( ext{H}_2 \)), which consists of two hydrogen atoms, we add together the atomic masses of each hydrogen atom. The atomic mass of a hydrogen atom is approximately 1.008 g/mol, so:
\[ ext{Molar mass of } ext{H}_2 = 2 imes 1.008 ext{ g/mol} = 2.016 ext{ g/mol}.\]
In our example, we first calculated the moles of hydrogen gas using its given mass and molar mass. Similarly, knowing the molar masses of iron (\( 55.845 ext{ g/mol} \)) and aluminum (\( 26.982 ext{ g/mol} \)) allows us to determine their respective masses when we know their molar quantities.

Chemical Reaction Equations

Chemical reaction equations are concise ways to represent what happens when substances react. Each equation reveals important information about the reactants, products, and their respective quantities involved in the reaction.
The reactions given in the exercise:

• Iron with hydrochloric acid: \( ext{Fe(s) + 2 HCl(aq) } \rightarrow ext{ FeCl}_2 ext{(aq) + H}_2 ext{(g)} \)
• Aluminum with hydrochloric acid: \( ext{2 Al(s) + 6 HCl(aq) } \rightarrow ext{ 2 AlCl}_3 ext{(aq) + 3 H}_2 ext{(g)} \)

These equations tell us that each metal reacts with hydrochloric acid to produce a chloride salt and hydrogen gas. The coefficients in the equation reflect the stoichiometry, ensuring that the number of atoms is balanced.
Interpreting these equations allows us to figure out exactly how much of each substance is needed and produced, enabling calculations for things like percent composition, as we calculated in the original task.