You are given 23.0 of aluminum and 28.0 of chlorine gas.
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 23.0 of aluminum?
Express your answer to three significant figures and include the appropriate units.
23.0 what Al? grams?
28.0 what Cl? grams?
2Al + 3Cl2 ==> 2AlCl3
mols Al = 23.0/atomic mass Al = ?
mols AlCl3 = the same (look at the coefficients).
To find the number of moles of aluminum chloride that can be produced, we need to determine the limiting reactant between aluminum and chlorine gas.
1. Start by finding the molar mass of aluminum (Al) and chlorine (Cl2):
- Aluminum (Al) has a molar mass of 26.98 g/mol.
- Chlorine gas (Cl2) has a molar mass of 70.91 g/mol.
2. Next, calculate the number of moles of aluminum (Al) using its given mass:
- Moles of aluminum (Al) = mass of aluminum (g) / molar mass of aluminum (g/mol)
- Moles of aluminum (Al) = 23.0 g / 26.98 g/mol
3. Determine the molar ratio between aluminum (Al) and aluminum chloride (AlCl3) from the balanced chemical equation:
- Balanced chemical equation: 2Al + 3Cl2 -> 2AlCl3
- The molar ratio between aluminum (Al) and aluminum chloride (AlCl3) is 2:2 or 1:1.
4. Since the ratio is 1:1, the number of moles of aluminum chloride will be the same as the number of moles of aluminum.
Hence, if you had excess chlorine, the number of moles of aluminum chloride that can be produced from 23.0 g of aluminum would be 23.0 / 26.98 = 0.852 moles. Round this to three significant figures, and the answer is 0.852 moles.