Based on this equation: 2AL + 3CuCl2 -> 2AlCl3 + Cu ... how many grams of copper(II) chloride dihydrate would be required to react completely with 1.20g of aluminum metal?
2Al + 3CuCl2 ➡ 2AlCl3 + 3Cu.
Moles of Al = 1.2/27 = 0.045 moles
2 moles of Al reacts with 3 moles of CuCl2.
Therefore, 1 mole of CuCl2 reacts with =2/3 x moles of Al
= 2/3 x 0.045
= 0.03 moles.
