Problem 16 - Chemistry and Chemical Reactivity

Question

Aluminum chloride, AlCl \(_{3}\), is made by treating scrap aluminum with chlorine. $$ 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{AlCl}_{3}(\mathrm{s}) $$ 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 AlC \(_{3}\) can be produced? (c) What mass of the excess reactant remains when the reaction is completed? (d) Set up an amounts table for this problem.

Step-by-Step Solution

Answer

(a) Cl_2 is the limiting reactant. (b) 5.07 g of AlCl_3 can be produced. (c) 1.67 g of Al remains. (d) Amounts table provided.

Step 1: Convert Masses to Moles

First, we need to convert the given masses of reactants to moles. \[ \text{Molar mass of Al} = 26.98 \text{ g/mol} \] \[ \text{Moles of Al} = \frac{2.70 \text{ g}}{26.98 \text{ g/mol}} = 0.100 \text{ mol} \] \[ \text{Molar mass of } Cl_2 = 70.90 \text{ g/mol} \] \[ \text{Moles of } Cl_2 = \frac{4.05 \text{ g}}{70.90 \text{ g/mol}} = 0.0571 \text{ mol} \]

Step 2: Determine the Limiting Reactant

Use the balanced chemical equation to find how many moles of each reactant are needed. The equation tells us that 2 moles of Al react with 3 moles of Cl_2. Calculate the theoretical amount required for a complete reaction: \[ \text{Al needed for } 0.0571 \text{ mol of } Cl_2 = \frac{2}{3} \times 0.0571 = 0.0381 \text{ mol of Al} \] Since we have 0.100 mol of Al available and need only 0.0381 mol, Cl_2 is the limiting reactant.

Step 3: Calculate Mass of AlCl3 Produced

The balanced equation shows that 2 moles of Al and 3 moles of Cl_2 produce 2 moles of AlCl_3. From Step 2, since Cl_2 is the limiting reactant and we have 0.0571 mol of Cl_2, the moles of AlCl_3 produced is also 0.0381 times 2/3: \[ \text{Moles of } AlCl_3 = 0.0381 \times \frac{2}{3} = 0.0381 \text{ mol } AlCl_3 \] The molar mass of AlCl_3 is \( 133.33 \text{ g/mol} \). Therefore, the mass of AlCl_3 is: \[ 0.0381 \times 133.33 = 5.07 \text{ g} \]

Step 4: Calculate Mass of Excess Al Remaining

We initially have 0.100 moles of Al, but only 0.0381 moles are required. The remaining moles of Al are: \[ 0.100 - 0.0381 = 0.0619 \text{ mol of Al} \] Convert the moles of excess Al back to mass: \[ 0.0619 \times 26.98 = 1.67 \text{ g of Al remaining} \]

Step 5: Set Up an Amounts Table

Create a table to summarize the initial, change, and final amounts of each substance in the reaction: | Reactant/Product | Initial Moles | Change in Moles | Final Moles | |------------------|---------------|-----------------|-------------| | Al | 0.100 | -0.0381 | 0.0619 | | Cl_2 | 0.0571 | -0.0571 | 0 | | AlCl_3 | 0 | +0.0381 | 0.0381 |

Key Concepts

Limiting Reactant

In stoichiometry, the limiting reactant is the substance in a chemical reaction that is completely consumed first, stopping the reaction from continuing. This means it determines the maximum amount of product that can be created. Identifying the limiting reactant involves comparing the mole ratios of the reactants to the stoichiometry of the balanced equation. For the given problem, aluminum (\(\mathrm{Al}\)) reacts with chlorine (\(\mathrm{Cl}_{2}\)), where the equation shows that 2 moles of \(\mathrm{Al}\) require 3 moles of \(\mathrm{Cl}_{2}\). By calculating the required amount of aluminum for the amount of chlorine provided versus what we have, we determined chlorine as the limiting reactant.

Molar Mass

Molar mass is a crucial concept in stoichiometry as it allows conversion between grams and moles, giving us a way to measure substances in manageable quantities. When solving problems, it's essential to know the molar mass of the substances involved. For instance, in this exercise, we need to know the molar masses of aluminum and chlorine. Aluminum (\(\mathrm{Al}\)) has a molar mass of 26.98 g/mol, and chlorine gas (\(\mathrm{Cl}_{2}\)) has a molar mass of 70.90 g/mol. Using these values, we calculate the number of moles of each substance by dividing the given mass by the molar mass (e.g., \(0.100 \, \text{moles of \(\mathrm{Al}\)}\) from \(2.70 \, \text{g}\)). This conversion is the first step to further stoichiometric calculations.

Chemical Reaction Quantities

In any chemical reaction, the quantities of reactants and products are interrelated through the balanced chemical equation. This equation dictates the proportions in which substances react and form products. For our given reaction \[ 2 \mathrm{Al}(\mathrm{s}) + 3 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{AlCl}_{3}(\mathrm{s}) \]2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride. This relationship allows us to calculate how much product can be formed from given amounts of reactants. Based on the quantity of chlorine available, we can calculate the aluminum chloride produced by maintaining these fixed ratios. This helps us predict the outcome of reactions and plan for consumption and yield.

Excess Reactant

In a chemical reaction, the excess reactant is the substance that is not completely used up during the process. It remains after the limiting reactant has been fully consumed. Identifying the excess reactant is important, as it helps determine both the efficiency of the reaction and waste management. From our exercise, after the chlorine is used up, aluminum is left over as the excess. Knowing that we start with \(0.100 \, \text{mol}\) of aluminum and only use \(0.0381 \, \text{mol}\), we calculate the remaining aluminum as \(0.0619 \, \text{mol}\). Converting this back to grams helps in finding the mass left after the reaction, which in this case is \(1.67 \, \text{g}\). This calculation shows how stoichiometric calculations guide resource utilization.