Question

a 54.0g sample of aluminum reacts with 508g of iodine.

Answer 1

$$2Al + 3I_2 = 2AlI_3$$

$$n(Al) = m/M = 54.0g/27.0g*mol^{-1} = 2mol$$

$$n(I_2) = m/M = 508.0g/254g*mol^{-1} = 2 mol$$

Aluminum in excess, calculation for iodine:

$$n(AlI_3) = 2/3*2 = 4/3 mol$$

$$m(AlI_3) = 4/3mol*408g*mol^{-1} = 544g$$