What is the maximum mass of aluminum chloride that can be formed when reacting 30.0g of aluminum with 35.0g of chlorine

To find the maximum mass of aluminum chloride that can be formed when reacting 30.0g of aluminum with 35.0g of chlorine, we need to determine the limiting reactant. The limiting reactant is the reactant that limits the amount of product formed in a chemical reaction.

To find the limiting reactant, we need to compare the amount of each reactant to the stoichiometry of the balanced chemical equation. The balanced chemical equation for the reaction between aluminum and chlorine is:

2Al + 3Cl2 -> 2AlCl3

According to the stoichiometry, 2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride.

First, let’s convert the masses of the reactants to moles using their molar masses. The molar mass of aluminum (Al) is 26.98 g/mol, and the molar mass of chlorine (Cl2) is 70.90 g/mol.

Number of moles of aluminum = Mass of aluminum / Molar mass of aluminum = 30.0 g / 26.98 g/mol = 1.112 mol

Number of moles of chlorine = Mass of chlorine / Molar mass of chlorine = 35.0 g / 70.90 g/mol = 0.493 mol

Now, let's compare the moles of each reactant with the stoichiometry of the balanced equation:

Aluminum: 1.112 mol (1.112 mol / 2 mol Al = 0.556)
Chlorine: 0.493 mol (0.493 mol / 3 mol Cl2 = 0.164)

From the calculations, it is clear that aluminum is in excess, and chlorine is the limiting reactant with 0.164 moles.

To find the maximum mass of aluminum chloride formed, we can use the stoichiometry of the balanced equation:

Molar mass of aluminum chloride (AlCl3) = 26.98 g/mol (Al) + 3 * 35.45 g/mol (Cl) = 133.34 g/mol

Number of moles of aluminum chloride formed = Number of moles of chlorine used (limiting reactant) = 0.164 mol

Mass of aluminum chloride formed = Number of moles of aluminum chloride * Molar mass of aluminum chloride
= 0.164 mol * 133.34 g/mol = 21.86 g

Therefore, the maximum mass of aluminum chloride that can be formed is 21.86 grams.

To determine the maximum mass of aluminum chloride that can be formed, we need to calculate the limiting reactant. This is the reactant that will be completely consumed in the reaction, determining the maximum amount of product that can be formed.

Let's calculate the number of moles of aluminum and chlorine:

1. Calculate the number of moles of aluminum:
Molecular mass of aluminum (Al) = 26.98 g/mol
Moles of aluminum = mass of aluminum / molecular mass of aluminum
= 30.0 g / 26.98 g/mol
≈ 1.11 mol

2. Calculate the number of moles of chlorine:
Molecular mass of chlorine (Cl₂) = 35.45 g/mol
Moles of chlorine = mass of chlorine / molecular mass of chlorine
= 35.0 g / 35.45 g/mol
≈ 0.99 mol

Next, we need to determine the stoichiometry of the reaction. From the balanced chemical equation for the reaction between aluminum and chlorine:

2 Al + 3 Cl₂ → 2 AlCl₃

We see that 2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride.

3. Determine the limiting reactant:
From the stoichiometry, we can calculate the number of moles of aluminum chloride formed by each reactant:
Moles of aluminum chloride formed from aluminum = 1.11 mol * (2 mol AlCl3 / 2 mol Al) = 1.11 mol
Moles of aluminum chloride formed from chlorine = 0.99 mol * (2 mol AlCl3 / 3 mol Cl₂) ≈ 0.66 mol

Since the moles of aluminum chloride formed from chlorine are smaller, chlorine is the limiting reactant.

Finally, we can calculate the maximum mass of aluminum chloride formed using the limiting reactant:

4. Calculate the mass of aluminum chloride:
Molecular mass of aluminum chloride (AlCl₃) = 133.34 g/mol
Mass of aluminum chloride = moles of aluminum chloride * molecular mass of aluminum chloride
≈ 0.66 mol * 133.34 g/mol
≈ 87.84 g

Therefore, the maximum mass of aluminum chloride that can be formed is approximately 87.84 grams.

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