Question

For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants. 2Al(s)+3Cl2(g)→2AlCl3(s)

Part B 7.4 g Al, 25.0 g Cl2 Express your answer using three significant figures. Part C 0.235 g Al, 1.20 g Cl2 Express your answer using three significant figures.

Answer 1

Answer 2

To determine the theoretical yield of the product, we need to compare the stoichiometry of the balanced equation with the given amounts of reactants. Let's calculate the theoretical yields for both scenarios:

Part B:

Given: Mass of Al = 7.4 g Mass of Cl2 = 25.0 g

We need to determine the limiting reactant first, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

1. Calculate the number of moles for each reactant: Moles of Al = mass / molar mass = 7.4 g / 26.98 g/mol = 0.274 mol Moles of Cl2 = mass / molar mass = 25.0 g / 70.90 g/mol = 0.353 mol

2. Determine the stoichiometric ratio between Al and Cl2: From the balanced equation, we can see that the stoichiometric ratio is 2:3 (2 moles of Al react with 3 moles of Cl2 to produce 2 moles of AlCl3).

3. Determine the limiting reactant: To determine the limiting reactant, we compare the moles of each reactant with the stoichiometric ratio.

• Al: 0.274 mol (actual) / 2 mol (stoichiometric ratio) = 0.137 (ratio)

• Cl2: 0.353 mol (actual) / 3 mol (stoichiometric ratio) = 0.118 (ratio)

The Cl2 has the smaller ratio, indicating that it is the limiting reactant.

4. Calculate the theoretical yield: From the stoichiometry, we know that 2 moles of AlCl3 are produced for every 3 moles of Cl2 reacted. Molar mass of AlCl3 = 133.34 g/mol (2 × 26.98 g/mol + 3 × 35.45 g/mol)

Theoretical yield = moles of limiting reactant (Cl2) × molar mass of AlCl3 Theoretical yield = 0.353 mol × 133.34 g/mol = 47.07 g

Therefore, the theoretical yield of AlCl3 is 47.07 g.

Part C:

Given: Mass of Al = 0.235 g Mass of Cl2 = 1.20 g

Follow the same steps as in Part B to determine the limiting reactant and calculate the theoretical yield.

1. Calculate the number of moles for each reactant: Moles of Al = 0.235 g / 26.98 g/mol = 0.00870 mol Moles of Cl2 = 1.20 g / 70.90 g/mol = 0.0169 mol

2. Determine the stoichiometric ratio between Al and Cl2 (from the balanced equation, it is 2:3).

3. Determine the limiting reactant:

• Al: 0.00870 mol (actual) / 2 mol (stoichiometric ratio) = 0.00435 (ratio)

• Cl2: 0.0169 mol (actual) / 3 mol (stoichiometric ratio) = 0.00563 (ratio)

The Al has the smaller ratio, indicating that it is the limiting reactant.

4. Calculate the theoretical yield: From the stoichiometry, we know that 2 moles of AlCl3 are produced for every 2 moles of Al reacted. Molar mass of AlCl3 = 133.34 g/mol

Theoretical yield = moles of limiting reactant (Al) × molar mass of AlCl3 Theoretical