For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants.

2Al(s) +3 Cl2(g) ---> 2 AlCl3(s)

A. 2.5gAl , 2.5g Cl2
B. 7.7 g Al , 25.2 g Cl2
C. 0.240 g Al , 1.10 g Cl2

Well, well, well! Let's calculate the theoretical yield for each scenario, shall we?

A. 2Al(s) + 3Cl2(g) ---> 2AlCl3(s)

First, let's see which reactant is the limiting one. To do that, we have to compare the moles of each reactant to the stoichiometric ratio in the balanced equation.

For 2.5 g of Al:
Molar mass of Al = 26.98 g/mol
Moles of Al = 2.5g / 26.98 g/mol = 0.093 mol

For 2.5 g of Cl2:
Molar mass of Cl2 = 2 x 35.45 g/mol = 70.9 g/mol
Moles of Cl2 = 2.5g / 70.9 g/mol = 0.035 mol

Now, let's compare the moles of each reactant to the stoichiometric ratio:

Al:Cl2 = 0.093 mol : 0.035 mol ≈ 2.7:1

Since the ratio is more than 2:3, we have an excess of Al. Therefore, Cl2 is the limiting reactant.

The molar ratio between Cl2 and AlCl3 is 3:2, so the theoretical yield of AlCl3 will be:

0.035 mol Cl2 x (2 mol AlCl3 / 3 mol Cl2) x (133.34 g AlCl3 / 1 mol AlCl3) ≈ 0.313 g AlCl3

B. 7.7 g Al, 25.2 g Cl2

For 7.7 g of Al:
Moles of Al = 7.7 g / 26.98 g/mol = 0.285 mol

For 25.2 g of Cl2:
Moles of Cl2 = 25.2 g / 70.9 g/mol = 0.355 mol

Al:Cl2 = 0.285 mol : 0.355 mol ≈ 0.8:1

In this case, the ratio is less than the stoichiometric ratio of 2:3, which means Al is the limiting reactant.

The molar ratio between Al and AlCl3 is 2:2, so the theoretical yield of AlCl3 will be:

0.285 mol Al x (2 mol AlCl3 / 2 mol Al) x (133.34 g AlCl3 / 1 mol AlCl3) = 95.44 g AlCl3

C. 0.240 g Al, 1.10 g Cl2

For 0.240 g of Al:
Moles of Al = 0.240 g / 26.98 g/mol ≈ 0.0089 mol

For 1.10 g of Cl2:
Moles of Cl2 = 1.10 g / 70.9 g/mol ≈ 0.0155 mol

Al:Cl2 = 0.0089 mol : 0.0155 mol ≈ 0.6:1

Once again, the ratio is less than 2:3, meaning Al is the limiting reactant.

0.0089 mol Al x (2 mol AlCl3 / 2 mol Al) x (133.34 g AlCl3 / 1 mol AlCl3) = 1.19 g AlCl3

So, in summary:
A. The theoretical yield of AlCl3 is 0.313 g.
B. The theoretical yield of AlCl3 is 95.44 g.
C. The theoretical yield of AlCl3 is 1.19 g.

Keep in mind that these values are calculated assuming a 100% yield. In reality, actual yields may vary. Now, isn't chemistry just a reaction worth laughing about?

To compute the theoretical yield of the product, you need to follow these steps:

1. Balance the chemical equation:

2Al(s) + 3Cl2(g) ---> 2AlCl3(s)

2. Calculate the molar mass of each reactant and product:

Molar mass of Al = 26.98 g/mol
Molar mass of Cl2 = 70.91 g/mol
Molar mass of AlCl3 = 133.34 g/mol

3. Convert the given mass of each reactant into moles:

moles of Al = mass of Al / molar mass of Al
moles of Cl2 = mass of Cl2 / molar mass of Cl2

4. Determine the limiting reactant:

The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. To identify the limiting reactant, you compare the mole ratio of reactants in the balanced equation to the actual mole ratio of reactants given.

Calculate the mole ratio of Al to Cl2 using the coefficients in the balanced equation (2:3).

mole ratio of Al to Cl2 = (2 moles of Al) / (3 moles of Cl2)

Now, compare this ratio to the actual mole ratio given by the question.

If the given mole ratio is less than the calculated mole ratio, then Al is the limiting reactant.
If the given mole ratio is greater than the calculated mole ratio, then Cl2 is the limiting reactant.

5. Once you determine the limiting reactant, use its moles in the balanced equation to calculate the moles of product that can be formed:

moles of AlCl3 = (moles of limiting reactant) x (2 moles of AlCl3 / 2 moles of Al)

6. Convert the moles of product into grams:

mass of AlCl3 = moles of AlCl3 x molar mass of AlCl3

Now, let's apply these steps to each of the given options:

A. 2.5 g Al and 2.5 g Cl2

Calculate the moles of Al and Cl2:
moles of Al = 2.5 g Al / 26.98 g/mol
moles of Cl2 = 2.5 g Cl2 / 70.91 g/mol

Determine the limiting reactant:
Compare the mole ratio of Al to Cl2:
mole ratio = (2/3)

The actual mole ratio is equal to the calculated mole ratio, so Al is the limiting reactant.

Calculate the moles of AlCl3 that can be formed:
moles of AlCl3 = (moles of limiting reactant) x (2 moles of AlCl3 / 2 moles of Al)
moles of AlCl3 = (moles of Al)

Convert the moles of AlCl3 to grams:
mass of AlCl3 = moles of AlCl3 x molar mass of AlCl3

B. 7.7 g Al and 25.2 g Cl2

Follow the same steps as before to calculate the theoretical yield.

C. 0.240 g Al and 1.10 g Cl2

Follow the same steps as before to calculate the theoretical yield.

Remember to perform all the calculations with the proper number of significant figures. Also, keep in mind that the theoretical yield assumes a perfect reaction with 100% yield, which may not be the case in real-world scenarios.

For (a).

1. Convert 2.5 g Al to moles.
2. Convert 2.5 g Cl2 to moles.
3a. Convert moles Al to moles AlCl3.
3b. Convert moles Cl2 to moles AlCl3.
3c. Compare 3a and 3b; the smaller number of moles of AlCl3 will tell you which material (Al or Cl2) is the limiting reagent.
4. Convert moles from 3c to grams.

The others are done the same way.