finding the largest mass of Aluminum chloride that can be formed when reacting with 31 grams of aluminum with 36 grams of chlorine
To determine the largest mass of Aluminum chloride that can be formed, we need to compare the amount of Aluminum and Chlorine to determine which reactant will be the limiting reagent. The limiting reagent is the one that will completely react and determines the amount of product formed.
1. Convert the given masses of Aluminum and Chlorine into moles:
- Aluminum (Al):
- Molar mass of Al = 26.98 g/mol
- Moles of Al = 31 g / 26.98 g/mol = 1.15 mol
- Chlorine (Cl₂):
- Molar mass of Cl₂ = 35.45 g/mol
- Moles of Cl₂ = 36 g / 35.45 g/mol = 1.02 mol
2. Write the balanced chemical equation for the reaction between Aluminum (Al) and Chlorine (Cl₂):
2 Al + 3 Cl₂ → 2 AlCl₃
3. Determine the stoichiometric ratio between Aluminum (Al) and Aluminum chloride (AlCl₃):
- From the balanced equation, we can see that 2 moles of Al react with 2 moles of AlCl₃.
4. Determine the stoichiometric ratio between Chlorine (Cl₂) and Aluminum chloride (AlCl₃):
- From the balanced equation, we can see that 3 moles of Cl₂ react with 2 moles of AlCl₃.
5. Compare the ratios of Aluminum (Al) and Chlorine (Cl₂) to the stoichiometric ratios in the balanced equation:
- Moles of Al / Stoichiometric ratio of Al = 1.15 mol / 2 = 0.57
- Moles of Cl₂ / Stoichiometric ratio of Cl₂ = 1.02 mol / 3 = 0.34
6. Identify the limiting reagent:
- The limiting reagent is Chlorine (Cl₂) since it is completely used up, and Aluminum is left over.
7. Use the limiting reagent to determine the maximum mass of Aluminum chloride (AlCl₃) formed:
- From the stoichiometry of the balanced equation, we can see that 3 moles of Cl₂ react to form 2 moles of AlCl₃.
- Moles of Cl₂ used = Moles of Cl₂ in excess - Moles of Cl₂ reacted
= 1.02 mol - (2 mol AlCl₃ / 3 mol Cl₂) × (1.02 mol) = 0.34 mol Cl₂
- Mass of AlCl₃ = Moles of AlCl₃ × Molar mass of AlCl₃
= (0.34 mol Cl₂ × (2 mol AlCl₃ / 3 mol Cl₂)) × (133.34 g/mol AlCl₃)
≈ 29 grams
Therefore, the largest mass of Aluminum chloride (AlCl₃) that can be formed is approximately 29 grams.
To find the largest mass of aluminum chloride that can be formed when reacting with 31 grams of aluminum and 36 grams of chlorine, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thereby limiting the amount of product that can be formed.
First, let's determine the moles of each reactant. To do this, we divide the given masses by their respective molar masses.
Molar mass of aluminum (Al) = 26.98 g/mol
Molar mass of chlorine (Cl₂) = 35.45 g/mol
Moles of aluminum (Al) = 31 g / 26.98 g/mol ≈ 1.15 mol
Moles of chlorine (Cl₂) = 36 g / 35.45 g/mol ≈ 1.02 mol
The balanced chemical equation for the reaction between aluminum and chlorine is:
2 Al + 3 Cl₂ → 2 AlCl₃
According to the stoichiometry of the balanced equation, it takes 2 moles of aluminum to react with 3 moles of chlorine to produce 2 moles of aluminum chloride. Therefore, the ratio of moles is 2:3 for aluminum to chlorine.
Now, let's compare the moles of each reactant to find the limiting reactant. Since the ratio of aluminum to chlorine is 2:3, we can multiply the moles of aluminum by the ratio to determine the expected number of moles of chlorine required to react completely.
Expected moles of chlorine = (2 moles of Al / 3 moles of Cl₂) * moles of Al ≈ (2/3) * 1.15 mol ≈ 0.77 mol
Since the moles of chlorine (1.02 mol) are greater than the expected moles (0.77 mol), chlorine is in excess and aluminum is the limiting reactant.
To find the mass of aluminum chloride formed, we need to use the stoichiometry of the balanced equation. The molar mass of aluminum chloride (AlCl₃) is 133.34 g/mol.
Molar mass of aluminum chloride (AlCl₃) = 133.34 g/mol
Moles of aluminum chloride (AlCl₃) = 1.15 mol (as aluminum is the limiting reactant)
Mass of aluminum chloride formed = moles of AlCl₃ * molar mass of AlCl₃
= 1.15 mol * 133.34 g/mol
≈ 146.82 grams
Therefore, the largest mass of aluminum chloride that can be formed is approximately 146.82 grams.