What mass of aluminum can be produced when 10.0g of aluminum chloride decomposes?
AlCl3.
mols AlCl3 = g/molar mass = 10/approx 138 = about 0.07
There is 1 mol Al in 1 mol AlCl3; therefore, you can get about 0.07 mols Al from 0.07 mols AlCl3. Then convert 0.07 mols Al to grams. g = mols x atomic mass = ?
All of my calculations should be repeated as I have estimated throughout.
To determine the mass of aluminum that can be produced when aluminum chloride decomposes, you need to use stoichiometry and the balanced equation of the reaction.
The balanced equation for the reaction where aluminum chloride decomposes to form aluminum is:
2AlCl₃ → 2Al + 3Cl₂
From the equation, we can see that 2 moles of aluminum chloride (AlCl₃) yield 2 moles of aluminum (Al). The molar mass of aluminum chloride is 133.34 g/mol, and the molar mass of aluminum is 26.98 g/mol.
Now let's calculate the number of moles of aluminum chloride (AlCl₃) in 10.0 g of the compound using the formula:
moles = mass / molar mass
moles of AlCl₃ = 10.0 g / 133.34 g/mol = 0.075 moles
Since the molar ratio between aluminum chloride (AlCl₃) and aluminum (Al) is 2:2, this means that the moles of aluminum produced (Al) will also be 0.075 moles.
Now, to determine the mass of aluminum (Al) produced, we will use the formula:
mass = moles × molar mass
mass of Al = 0.075 moles × 26.98 g/mol = 2.02 g
Therefore, when 10.0 g of aluminum chloride decomposes, approximately 2.02 g of aluminum can be produced.