Answer:
Hence, 0.455 mol of aluminum chloride could be produced from 12.0 grams of aluminum.
Explanation:
Given data:
•Mass of aluminum is 12.0 grams.
To find:
•Number of moles of aluminum chloride produced from Al.
The given reaction is as follows:
2A1+ 3Cl₂ → 2AlCl3
First, we have to find the number of moles of aluminum by the relation given below:
n = m M;
Where:
•n is the number of moles.
•m is the mass in grams.
•M is the molar mass in g/mol.
The molar mass of Al is 26.98 g/mol. Substitute the values in above equation to obtain the number of moles of aluminium as follows:
n = (12.0g)/(26.98g /mol)
= 0.445 mol
From the reaction stoichiometry, we can see that 2 moles of Al forms 2 moles of aluminium chloride (A1C13). This means, the number of moles of Al and AlCl3 is equal.
So, 0.445 mol Al will give 0.455 mol AlC13.
