Question

How many grams of Al is needed to make 0.59g of AlCl3?

Answer 1

2Al(s)+3Cl2(g)→2AlCl3(s)

Moles of AlCl3$=\frac{0.59}{133}=4.4×10^{-3}$

Moles of Al $=4.4×10^{-3}$

grams of Al = $4.4×10^{-3}×27 =0.12g$