Calculate the theoretical amount of aluminum iodide that should have been produced in this lab.
We used .25 g of Al and 4 g of I. I have the balanced equation figured out, it's 3 I2 + 2 Al -> 2 AlI3. Now what do I do?
We used .25 g of Al and 4 g of I. I have the balanced equation figured out, it's 3 I2 + 2 Al -> 2 AlI3. Now what do I do?
-
3 I2 + 2 Al → 2 AlI3
(0.25 g Al) / (26.9815 g Al/mol) = 0.0092656 mol Al
(4 g I2) / (253.8089 g I2/mol) = 0.015760 mol I2
0.0092656 mole of Al would react completely with 0.0092656 x (3/2) = 0.013898 mole of I2, but there is more I2 present than that, so I2 is in excess and Al is the limiting reactant.
(0.0092656 mol Al) x (2/2) x (407.694958 g AlI3/mol) = 3.8 g AlI3 in theory
(0.25 g Al) / (26.9815 g Al/mol) = 0.0092656 mol Al
(4 g I2) / (253.8089 g I2/mol) = 0.015760 mol I2
0.0092656 mole of Al would react completely with 0.0092656 x (3/2) = 0.013898 mole of I2, but there is more I2 present than that, so I2 is in excess and Al is the limiting reactant.
(0.0092656 mol Al) x (2/2) x (407.694958 g AlI3/mol) = 3.8 g AlI3 in theory