Problem 24 - Chemistry and Chemical Reactivity

Question

Aluminum chloride, AlCl_, is made by treating scrap eluminum with chlorine. $$ 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{s}) $$ If you begin with \(2.70 \mathrm{g}\) of \(\mathrm{Al}\) and \(4.05 \mathrm{g}\) of \(\mathrm{Cl}_{2}\) (a) Which reactant is limiting? (b) What mass of AlCl \(_{3}\) can be produced? (c) What mass of the excess reactant remains when the reaction is completed?

Step-by-Step Solution

Answer

(a) Cl\(_2\) is limiting. (b) 5.08 g of AlCl\(_3\) is produced. (c) 1.67 g of Al remains.

Step 1: Calculate Moles of Al and Cl2

First, we need to find the moles of Al and Cl\(_2\). Use the molar mass of Al (\(26.98\text{ g/mol}\)) and Cl\(_2\) (\(70.90\text{ g/mol}\)).\[\text{Moles of Al} = \frac{2.70 \text{ g}}{26.98 \text{ g/mol}} \approx 0.1000 \text{ mol}\]\[\text{Moles of Cl}_2 = \frac{4.05 \text{ g}}{70.90 \text{ g/mol}} \approx 0.0571 \text{ mol}\]

Step 2: Determine Limiting Reactant

From the balanced equation, the reaction uses a 2:3 ratio of Al to Cl\(_2\). This means 2 mol of Al react with 3 mol of Cl\(_2\).Calculate the required moles of Cl\(_2\) needed to react with 0.1000 mol of Al:\[\text{Required Cl}_2 = 0.1000 \text{ mol Al} \times \frac{3 \text{ mol Cl}_2}{2 \text{ mol Al}} = 0.1500 \text{ mol Cl}_2\]Since we have only 0.0571 mol of Cl\(_2\), Cl\(_2\) is the limiting reactant.

Step 3: Calculate Mass of AlCl3 Produced

Use the limiting reactant (Cl\(_2\)) to determine the amount of AlCl\(_3\) produced. Each 3 mol of Cl\(_2\) produces 2 mol of AlCl\(_3\).Calculate the moles of AlCl\(_3\):\[\text{Moles of AlCl}_3 = 0.0571 \text{ mol Cl}_2 \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} = 0.0381 \text{ mol AlCl}_3\]Then find the mass of AlCl\(_3\). The molar mass of AlCl\(_3\) is \(133.33 \text{ g/mol}\):\[\text{Mass of AlCl}_3 = 0.0381 \text{ mol} \times 133.33 \text{ g/mol} \approx 5.08 \text{ g}\]

Step 4: Calculate Remaining Excess Reactant (Al)

Now, calculate how much Al is used.From Step 2, Cl\(_2\) is limiting and 0.0571 mol Cl\(_2\) is fully consumed. We react these with Al in a 3:2 ratio:\[\text{Moles of Al used} = 0.0571 \text{ mol Cl}_2 \times \frac{2 \text{ mol Al}}{3 \text{ mol Cl}_2} = 0.0381 \text{ mol Al}\]To find the remaining Al, calculate initial - used:\[\text{Initial moles of Al} - \text{Moles of Al used} = 0.1000 \text{ mol} - 0.0381 \text{ mol} = 0.0619 \text{ mol}\]Convert remaining moles back to grams:\[0.0619 \text{ mol} \times 26.98 \text{ g/mol} \approx 1.67 \text{ g}\]

Step 5: Conclusion

The limiting reactant is Cl\(_2\). The mass of AlCl\(_3\) that can be produced is approximately 5.08 g. There are approximately 1.67 g of Al remaining unreacted.

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely used up first. This limits the amount of products that can be formed because once it is consumed, the reaction stops. To identify the limiting reactant, you first need to determine the amount of product each reactant could produce if it were completely used. Compare these amounts to see which one is smaller.

In the given exercise, we have aluminum (Al) and chlorine gas (\(\text{Cl}_2\)). The balanced equation \(2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3\) shows that 2 moles of aluminum react with 3 moles of chlorine. Start by converting the given masses of reactants into moles: \(0.1000 \text{ mol of Al}\) and \(0.0571 \text{ mol of Cl}_2\).

The stoichiometric ratio requires \(0.1500 \text{ mol}\) of \(\text{Cl}_2\) to fully react with \(0.1000 \text{ mol of Al}\). But we only have \(0.0571 \text{ mol of Cl}_2\) available, making \(\text{Cl}_2\) the limiting reactant because it runs out first.

Understanding the limiting reactant concept is crucial for correctly predicting the amount of product formed and for knowing when a reaction will complete.

Mole Calculations

Mole calculations are the backbone of converting mass into more usable chemical quantities. A "mole" is a unit that counts particles, but instead of counting atoms or molecules one by one, which is impractical due to their size, chemists use the "mole." A mole equals Avogadro's number, which is \(6.022 \times 10^{23}\) particles. This makes calculations involving moles much easier! To convert from grams to moles, you divide the mass of a substance by its molar mass (grams per mole).

Here's how it works in the problem: The mass of aluminum is \(2.70 \text{ g}\) and the mass of chlorine is \(4.05 \text{ g}\). Using the molar masses provided—\(26.98 \text{ g/mol}\) for Al and \(70.90 \text{ g/mol}\) for \(\text{Cl}_2\)—the moles of each reactant are calculated as follows:

• Moles of Al: \(\frac{2.70 \text{ g}}{26.98 \text{ g/mol}} \approx 0.1000 \text{ mol}\)
• Moles of \(\text{Cl}_2\): \(\frac{4.05 \text{ g}}{70.90 \text{ g/mol}} \approx 0.0571 \text{ mol}\)

Understanding how to perform mole calculations ensures you can determine how much of each substance participates in the reaction and predicts the outcome.

Balanced Chemical Equations

Balanced chemical equations are essential for understanding chemical reactions. They show the exact numbers of molecules or moles of reactants and products involved in a reaction. Balancing an equation ensures that the same number of each type of atom appears on both sides of the equation, respecting the Law of Conservation of Mass.In our exercise, the balanced chemical equation is: \(2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3\).

Notice each element is balanced on both sides:

• Aluminum: 2 atoms on both sides
• Chlorine: \(6\) atoms on both sides

Balanced equations allow for the accurate calculation of reactants needed and products formed. They also help predict the amount of each substance remaining at the end of a reaction. For every \(3 \text{Cl}_2\) molecules, only \(2 \text{AlCl}_3\) molecules are formed, making the stoichiometric ratios crucial to solve problems related to reactions.

When practicing stoichiometry, always begin with a balanced equation to ensure that you base further calculations on accurate scientific principles.