Answer:
10.3 g Al
Explanation:
To find the excess mass of Al, you need to (1) convert grams I₂ to moles I₂ (via molar mass), then (2) convert moles I₂ to moles AlI₃ (via mole-to-mole ratio from equation coefficients), then (3) convert moles AlI₃ to grams AlI₃ (via molar mass). Now that you have the actual amount of AlI₃ produced, you need to (4) convert grams AlI₃ to moles AlI₃ (via molar mass), then (5) convert moles AlI₃ to moles Al (via mole-to-mole ratio from equation coefficients), and then (6) convert moles Al to grams Al (via molar mass). Now that you know the amount of Al actually needed to produce the product, you need to (7) find the excess mass of Al.
Molar Mass (Al): 26.982 g/mol
Molar Mass (I₂): 2(126.90 g/mol)
Molar Mass (I₂): 253.8 g/mol
Molar Mass (AlI₃): 26.982 g/mol + 3(126.90 g/mol)
Molar Mass (AlI₃): 407.682 g/mol
2 Al(s) + 3 I₂(g) -------> 2 AlI₃(g)
113 g I₂ 1 mole 2 moles AlI₃ 407.682 g
------------- x ---------------- x ---------------------- x ------------------- = 121 g AlI₃
253.8 g 3 moles I₂ 1 mole
121 g AlI₃ 1 mole 2 moles Al 26.982 g
--------------- x ------------------ x --------------------- x ----------------- = 8.01 g Al
407.682 g 2 moles AlI₃ 1 mole
Starting Amount - Mass Needed = Excess
18.3 g Al - 8.01 g Al = 10.3 g Al
