Answer:
596 g of AlCl₃
Explanation:
The reaction is:
2Al + 3Cl₂ → 2AlCl₃
Firstly we need to determine the limiting reactant and we need to determine the moles of each
135 g . 1 mol / 26.98 g = 5 moles of Al
475 g . 1mol / 70.9 g = 6.7 moles of chlorine
Ratio in the reaction is 3:2
3 moles of chlorine gas need 2 moles of Al to react
Then, our 6.7 moles of gas might react to (6.7 . 2 ) /3 = 4.47 moles of Al
We have 5 moles of Al and we need 4.47 moles, so this is the excess reactant. In conclussion, Cl₂ is the limiting.
2 moles of Al react to 3 moles of gas
Our 5 moles of Al may react to (5 . 3 )/2 = 7.5 moles of gas
It's ok because we only have 6.7 moles, and there is not enough gas.
Now we calculate the product's mass
3 moles of gas can produce 2 moles of salt
Then, our 6.7 moles of gas will produce (6.7 . 2) /3 = 4.47 moles
We convert moles to mass: 4.47 mol . 133.33 g/mol = 596 g of AlCl₃
