2Al(s)+3Cl2 = 2AlCl3
Limiting Reactant is Cl with 0.329mol.
What is the maximum mass of aluminum chloride that can be formed when reacting 30.0g of aluminum with 35.0g of chlorine?
35g of Al*(1mole of Cl/35.45g)= mole of Cl
3 moles of Cl= 2 moles of AlCl3
moles of Cl *(2 moles of AlCl3/3 moles of Cl)= moles of AlCl3
moles of ALCl3 *(133.34 g of AlCl3/mol)= mass of AlCl3
To calculate the maximum mass of aluminum chloride that can be formed, we need to determine the limiting reactant first.
First, let's calculate the number of moles of aluminum (Al) and chlorine (Cl2) using their respective molar masses.
Molar mass of Al: 26.98 g/mol
Molar mass of Cl2: 2 * (35.45 g/mol) = 70.90 g/mol
Number of moles of Al:
30.0 g Al * (1 mol Al / 26.98 g Al) = 1.11 mol Al
Number of moles of Cl2:
35.0 g Cl2 * (1 mol Cl2 / 70.90 g Cl2) = 0.493 mol Cl2
According to the balanced chemical equation, the stoichiometric ratio of Al to Cl2 is 2:3. Therefore, for every 2 moles of Al, we need 3 moles of Cl2.
Let's calculate the maximum moles of AlCl3 that can be formed using the limiting reactant.
Based on the ratio, for every 3 moles of Cl2, we can form 2 moles of AlCl3.
0.329 mol Cl2 * (2 mol AlCl3 / 3 mol Cl2) = 0.219 mol AlCl3
Finally, let's calculate the maximum mass of AlCl3 formed by multiplying moles by the molar mass of AlCl3.
Molar mass of AlCl3: 26.98 g/mol (Al) + 3 * (35.45 g/mol) (Cl) = 133.34 g/mol
Mass of AlCl3:
0.219 mol AlCl3 * (133.34 g AlCl3 / mol) = 29.2 g AlCl3
Therefore, the maximum mass of aluminum chloride that can be formed when reacting 30.0 g of aluminum with 35.0 g of chlorine is 29.2 grams.
To find the maximum mass of aluminum chloride that can be formed, we need to determine the limiting reactant first.
1. Start by finding the molar masses of aluminum (Al) and chlorine (Cl2):
- The molar mass of Al = 26.98 g/mol
- The molar mass of Cl2 = 2 x 35.45 g/mol = 70.90 g/mol
2. Convert the masses of aluminum and chlorine to moles:
- Moles of aluminum = mass (g) / molar mass (g/mol)
Moles of aluminum = 30.0 g / 26.98 g/mol = 1.112 mol
- Moles of chlorine = mass (g) / molar mass (g/mol)
Moles of chlorine = 35.0 g / 70.90 g/mol = 0.493 mol
3. Determine the molar ratio between aluminum and chlorine in the balanced chemical equation:
According to the balanced equation, 2 moles of aluminum react with 3 moles of chlorine to form 2 moles of aluminum chloride.
If we divide the number of moles of chlorine by its coefficient in the balanced equation, we get a ratio:
Moles of chlorine / coefficient of chlorine = 0.493 mol / 3 = 0.1643 mol Cl2
Since there is a 1:3 molar ratio between aluminum and chlorine, we can multiply the moles of chlorine by 2/3 to find the expected moles of aluminum.
Moles of aluminum = Moles of chlorine x 2/3 = 0.1643 mol x (2/3) = 0.1095 mol Al
4. Compare the calculated moles of aluminum and chlorine to determine the limiting reactant:
Since the moles of chlorine is less than the moles of aluminum, chlorine is the limiting reactant.
5. Next, determine the maximum moles of aluminum chloride that can be formed based on the limiting reactant:
According to the balanced equation, 3 moles of chlorine react to produce 2 moles of aluminum chloride.
Thus, the expected moles of aluminum chloride formed from 0.1643 mol of chlorine are:
Moles of aluminum chloride = Moles of chlorine x (2/3) x (2/3) = 0.1643 mol x (2/3) x (2/3) = 0.1095 mol AlCl3
6. Finally, calculate the maximum mass of aluminum chloride using the moles and molar mass of AlCl3:
Maximum mass of aluminum chloride = Moles of aluminum chloride x molar mass of AlCl3
Maximum mass of aluminum chloride = 0.1095 mol x 133.34 g/mol = 14.6 g (rounded to three significant figures)
Therefore, the maximum mass of aluminum chloride that can be formed when reacting 30.0 g of aluminum with 35.0 g of chlorine is approximately 14.6 g.