Question

if you react 5.00g of aluminum with 20.0g of iodine, how much aluminum iodide do you expect to produce from this reaction? what is the limiting reactant ? what is the excess reactant? how much of the excess reactant do you expect to have left over?

Answer 1

2Al + 3I2 >>>2AlI₃

so for each two moles of aluminum, one needs three moles of I2.
you have 5/27= .185 moles of Al
so you need then 3/2 *.185 moles of I2, or
1.5*.185*127*2=70.5 grams. You have far less than that, so Iodine is the limiting reageant.
How much can you expedt:
moles of I2=20/(2*127)=.079 moles, so you should get 2/3*.079 moles AlI3.