Problem 9 - Chemistry and Chemical Reactivity

Question

Like many metals, aluminum reacts with a halogen to give a metal halide (see Figure 3.1 ). $$ 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Br}_{2}(\ell) \longrightarrow \mathrm{Al}_{2} \mathrm{Br}_{6}(\mathrm{s}) $$ What mass of \(\mathrm{Br}_{2}\), in grams, is required for complete reaction with \(2.56 \mathrm{g}\) of Al? What mass of white, solid \(\mathrm{Al}_{2} \mathrm{Br}_{6}\) is expected?

Step-by-Step Solution

Answer

22.7 g of \(\mathrm{Br}_{2}\) and 25.3 g of \(\mathrm{Al}_{2}\mathrm{Br}_{6}\) are needed and expected, respectively.

Step 1: Determine Molar Mass of Al

The molar mass of aluminum (Al) is approximately \(26.98 \text{ g/mol}\). This information will be used to convert the given mass of Al to moles.

Step 2: Convert Mass of Al to Moles

Given \(2.56 \text{ g}\) of Al, we convert this mass to moles using the formula \( \, \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{2.56}{26.98} \, \text{mol} \approx 0.0948 \, \text{mol}\).

Step 3: Use Stoichiometry to Determine Moles of \(\mathrm{Br}_{2}\)

According to the balanced equation, \(2 \text{ mol of Al}\) reacts with \(3 \text{ mol of } \mathrm{Br}_{2}\). Thus, \( 0.0948 \text{ mol of Al}\) will require \( \frac{3}{2} \times 0.0948 \, \text{mol} \approx 0.142 \text{ mol of } \mathrm{Br}_{2}\).

Step 4: Determine Molar Mass of \(\mathrm{Br}_{2}\)

The molar mass of \(\mathrm{Br}_{2}\) is approximately \(79.9 \, \text{g/mol} \times 2 = 159.8 \, \text{g/mol}\).

Step 5: Convert Moles of \(\mathrm{Br}_{2}\) to Mass

Using the formula \( \, \text{mass of } \mathrm{Br}_{2} = \text{moles of } \mathrm{Br}_{2} \times \text{molar mass of } \mathrm{Br}_{2} = 0.142 \times 159.8 \, \text{g} \approx 22.7 \, \text{g}\) of \(\mathrm{Br}_{2}\) is required.

Step 6: Use Stoichiometry to Determine Moles of \(\mathrm{Al}_{2}\mathrm{Br}_{6}\)

According to the balanced equation, \(2 \text{ mol of Al}\) will produce \(1 \text{ mol of } \mathrm{Al}_{2}\mathrm{Br}_{6}\). Thus, \(0.0948 \text{ mol of Al}\) will produce \(0.0474 \text{ mol of } \mathrm{Al}_{2}\mathrm{Br}_{6}\).

Step 7: Determine Molar Mass of \(\mathrm{Al}_{2}\mathrm{Br}_{6}\)

The molar mass of \(\mathrm{Al}_{2}\mathrm{Br}_{6}\) is approximately \((2 \times 26.98) + (6 \times 79.9) = 533.4 \text{ g/mol}\).

Step 8: Convert Moles of \(\mathrm{Al}_{2}\mathrm{Br}_{6}\) to Mass

Using the formula \( \, \text{mass of } \mathrm{Al}_{2}\mathrm{Br}_{6} = \text{moles of } \mathrm{Al}_{2}\mathrm{Br}_{6} \times \text{molar mass of } \mathrm{Al}_{2}\mathrm{Br}_{6} = 0.0474 \times 533.4 \, \text{g} \approx 25.3 \, \text{g}\) of \(\mathrm{Al}_{2}\mathrm{Br}_{6}\) is expected.

Key Concepts

Molar Mass Calculation

Molar mass is an essential concept in stoichiometry as it helps us calculate the amount of substance present in a given mass. It represents the mass of one mole of a substance, expressed in grams per mole (g/mol). This value is obtained by adding together the average atomic masses of all the atoms present in a molecule.
For instance, when calculating the molar mass of the compound \( \text{Br}_2 \), we look at the periodic table to find the atomic mass of a bromine atom, approximately \( 79.9 \) g/mol. Since the bromine molecule \( \text{Br}_2 \) contains two bromine atoms, the molar mass is doubled, resulting in \( 159.8 \) g/mol.
Similarly, the molar mass of aluminum \( \text{Al} \) is \( 26.98 \) g/mol. These calculations allow us to convert between grams and moles, providing a bridge for further stoichiometric analysis.

Balanced Chemical Equation

Understanding a balanced chemical equation is pivotal in stoichiometry. It represents a chemical reaction with the same number of each type of atom on both sides of the equation, ensuring the conservation of mass.
For the reaction between aluminum and bromine to form aluminum bromide, the balanced equation is:

• \( 2 \text{Al} + 3 \text{Br}_2 \rightarrow \text{Al}_2\text{Br}_6 \)

This equation tells us that two moles of aluminum (\( \text{Al} \)) react with three moles of bromine (\( \text{Br}_2 \)) to produce one mole of aluminum bromide (\( \text{Al}_2\text{Br}_6 \)). Balancing equations is a crucial skill, as it forms the basis for determining the quantitative relationships between reactants and products.
Additionally, the balanced equation directly informs us of how much of each reactant is needed and how much product we can expect to form in a chemical reaction.

Mass-to-Mole Conversion

Converting between mass and moles is a fundamental skill in stoichiometry used to understand the quantities involved in chemical reactions. This conversion uses the relationship between mass and molar mass.
If we have a given mass of a substance and its molar mass, we can determine the number of moles present using the formula:

• Moles = \( \frac{\text{mass}}{\text{molar mass}} \)

For example, if we have \( 2.56 \) grams of aluminum (\( \text{Al} \)) and its molar mass is \( 26.98 \) g/mol, we calculate the moles of aluminum as approximately \( \frac{2.56}{26.98} \approx 0.0948 \) moles.
This step is critical for further stoichiometric calculations, allowing us to compare the amount of different substances involved in the reaction according to the balanced chemical equation.