If 0.0490 mol of solid Al reacts stoichiometrically according to the balanced equation with 990 mL of aqueous Cl-, what molarity (M) of Cl- is required?

Question

3 CuCl2(aq) + 2 Al(s) → 3 Cu + 2AlCl3(aq)

Answer 1

You have 0.0490 mole Al.
I assume that is 990 mL aq CuCl2. Moles CuCl2 needed = moles Al x (3 moles CuCl2/2 moles Al) = ??
M CuCl2 = moles/L = ?? moles/0.990 L.
The Cl^- must be 2x that. Check my thinking.

Answer 2

i did 0.049x2/3=mole of CuCl2
then mole of cucl2/.99 then times by 2 to get 0.0660 but its wrong, the actual answer is 0.148. Can you double check this again why i did it wrong

Answer 3

You messed up on the conversion. Look at my response again. It is 0.049 x (3/2) and that x 2 and NOT 0.049 x (2/3) and that x 2. ;-).
I don't get 0.148 but 0.1470000000000.

Answer 4

Sorry about that. I forgot to divide by 0.99 and the answer is 0.148. But your error still is the same. Multiply by 3/2 and not by 2/3.