how many grams of AlCl3 is produced if 15.5 grams of Al are reacted with excess Cl2 gas? 2Al + 3Cl2 > 2AlCl3
mols Al = grams/atomic mass Al
Using the coefficients in the balanced equation, convert mols Al to mols AlCl3.
Now convert mols AlCl3 to grams. g AlCl3 = mols AlCl3 x molar mass AlCl3.
To determine how many grams of AlCl3 are produced, we need to use stoichiometry and the balanced chemical equation.
First, let's find the molar mass of Al and AlCl3:
- The molar mass of Al (aluminum) is 26.98 g/mol.
- The molar mass of AlCl3 (aluminum chloride) is 133.34 g/mol (3 times the molar mass of Cl + the molar mass of Al).
Next, we'll calculate the number of moles of Al we have:
Moles of Al = mass of Al / molar mass of Al
Moles of Al = 15.5 g / 26.98 g/mol ≈ 0.574 moles of Al
From the balanced equation, we can see that the ratio of moles of Al to moles of AlCl3 is 2:2. So, the moles of AlCl3 produced will also be 0.574 moles.
Finally, we'll convert moles of AlCl3 to grams of AlCl3:
Grams of AlCl3 = moles of AlCl3 x molar mass of AlCl3
Grams of AlCl3 = 0.574 moles x 133.34 g/mol ≈ 76.52 grams of AlCl3
Therefore, approximately 76.52 grams of AlCl3 are produced when 15.5 grams of Al are reacted with excess Cl2 gas.