3 CaSO4 + 2 AlCl3 Al2(SO4)3 + 3 CaCl2
How many moles of aluminum chloride are required to produce 97.0 g of calcuim chloride in the above reaction?
To determine how many moles of aluminum chloride are required to produce 97.0 g of calcium chloride, we need to use the balanced chemical equation and the molar masses of the compounds involved.
First, let's calculate the molar mass of calcium chloride (CaCl2):
- The molar mass of calcium (Ca) is 40.08 g/mol.
- The molar mass of chlorine (Cl) is 35.45 g/mol (since there are two chlorine atoms in CaCl2, we multiply the molar mass of Cl by 2).
- Adding the molar masses together, we get: 40.08 g/mol + (35.45 g/mol × 2) = 40.08 g/mol + 70.9 g/mol = 110.98 g/mol.
Now that we know the molar mass of calcium chloride, we can calculate the number of moles using the formula:
moles = mass / molar mass.
moles of calcium chloride = 97.0 g / 110.98 g/mol = 0.872 mol.
Looking at the balanced chemical equation:
3 CaSO4 + 2 AlCl3 → Al2(SO4)3 + 3 CaCl2.
We can see that for every 2 moles of aluminum chloride (AlCl3), we produce 3 moles of calcium chloride (CaCl2). Therefore, there is a stoichiometric ratio of 2:3 between aluminum chloride and calcium chloride.
To find the number of moles of aluminum chloride required, we can set up a proportion using the stoichiometric ratio:
(2 mol AlCl3 / 3 mol CaCl2) = (x mol AlCl3 / 0.872 mol CaCl2).
Solving for x, we find:
x = (2/3) × 0.872 mol = 0.581 mol.
Therefore, approximately 0.581 moles of aluminum chloride are required to produce 97.0 g of calcium chloride in the given reaction.