Problem 3 - Chemistry and Chemical Reactivity

Question

Like many metals, aluminum reacts with a halogen (here the orange-brown liquid \(\mathrm{Br}_{2}\) ) to give a metal halide, aluminum bromide. (The white solid on the lip of the beaker at the end of the reaction is \(\mathrm{Al}_{2} \mathrm{Br}_{6} .\) ) $$2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Br}_{2}(\ell) \rightarrow \mathrm{Al}_{2} \mathrm{Br}_{6}(\mathrm{s})$$ What mass of \(\mathrm{Br}_{2}\), in grams, is required for complete reaction with \(2.56 \mathrm{g}\) of \(\mathrm{Al}\) ? What mass of white, solid \(\mathrm{Al}_{2} \mathrm{Br}_{6}\) is expected?

Step-by-Step Solution

Answer

22.75 g of \( \text{Br}_2 \) is required, and 25.31 g of \( \text{Al}_2\text{Br}_6 \) is expected.

Step 1: Determine Molar Mass of Reactants and Products

First, calculate the molar mass of aluminum (\( \text{Al} \)), bromine (\( \text{Br}_2 \)), and aluminum bromide (\( \text{Al}_2\text{Br}_6 \)). The molar mass of \( \text{Al} \) is approximately 26.98 g/mol. The molar mass of \( \text{Br}_2 \) is approximately 79.90 g/mol each, so \( \text{Br}_2 \) is about 159.80 g/mol. The molar mass of \( \text{Al}_2\text{Br}_6 \) is calculated as \( 2\times26.98 + 6\times79.90 \approx 533.34 \) g/mol.

Step 2: Calculate Moles of Aluminum

Given that 2.56 g of \( \text{Al} \) is used, calculate the number of moles: \[ \text{moles of Al} = \frac{2.56\, \text{g}}{26.98\, \text{g/mol}} \approx 0.0949\, \text{mol} \].

Step 3: Relate Moles of Aluminum to Bromine

From the balanced chemical equation: \( 2\, \text{mol of Al} \) reacts with \( 3\, \text{mol of Br}_2 \). Therefore, \( 0.0949\, \text{mol of Al} \) will react with \[ \frac{3}{2} \times 0.0949\, \text{mol of Br}_2 \approx 0.1424\, \text{mol of Br}_2 \].

Step 4: Calculate Mass of Bromine Required

Convert moles of bromine to grams using its molar mass: \[ \text{mass of } \text{Br}_2 = 0.1424\, \text{mol} \times 159.80\, \text{g/mol} \approx 22.75\, \text{g} \].

Step 5: Calculate Expected Mass of Aluminum Bromide

Using the reaction, \( 2\, \text{mol Al} \rightarrow 1\, \text{mol Al}_2\text{Br}_6 \). Thus, \( 0.0949\, \text{mol of Al} \) produces \( \frac{1}{2} \times 0.0949 = 0.04745\, \text{mol of Al}_2\text{Br}_6 \). Convert this to grams: \[ \text{mass of Al}_2\text{Br}_6 = 0.04745\, \text{mol} \times 533.34\, \text{g/mol} \approx 25.31\, \text{g} \].

Key Concepts

Understanding Molar Mass

Molar mass is a crucial concept in chemistry because it links the amount of a substance (in moles) to its mass (in grams). Each element has a molar mass equivalent to its atomic weight on the periodic table, expressed in grams per mole (g/mol).

For aluminum (Al), the molar mass is approximately 26.98 g/mol. Bromine (\(\text{Br}_2\)) has a molecular molar mass of approximately 159.80 g/mol since it's diatomic. This value results from each bromine atom having a molar mass of about 79.90 g/mol.

To find the molar mass of complex compounds like aluminum bromide (Al\(_2\)Br\(_6\)), we sum the molar masses of all atoms in the formula. Hence, \(2 \times 26.98 + 6 \times 79.90 = 533.34\) g/mol. Knowing these values helps in converting between mass and moles, essential for calculations in stoichiometry.

Basics of Chemical Reactions

Chemical reactions involve the rearrangement of atoms to form new substances. The balanced chemical equation depicts how reactants transform into products.

The equation for the synthesis of aluminum bromide is:\[2 \text{Al}(\text{s}) + 3 \text{Br}_2(\ell) \rightarrow \text{Al}_2\text{Br}_6(\text{s})\]This equation tells us that 2 moles of aluminum react with 3 moles of bromine to form 1 mole of aluminum bromide.

Balancing equations ensures we account for the conservation of mass, meaning the total mass and number of atoms on each side must equal each other. It also provides the mole ratio for stoichiometric calculations, critical for determining how much of a reactant is needed to form a certain amount of product.

Aluminum Bromide Synthesis

The synthesis of aluminum bromide, a common chemical reaction in stoichiometry, illustrates how metals can react with halogens to form metal halides. In this reaction, pieces of solid aluminum are combined with liquid bromine to produce solid aluminum bromide.

This process requires understanding both the reactants and the product. With a known mass of aluminum (e.g., 2.56 g as in the exercise), you can calculate the moles of aluminum using its molar mass. From the chemical equation, find the moles of bromine required by using the mole ratio (2:3 in this case).

To find out how much bromine is needed in grams, convert the moles of bromine to grams using its molar mass. Finally, using the balanced equation, calculate the mass of aluminum bromide formed by using the number of moles and its molar mass, resulting in a complete stoichiometric conversion.