Question

Suppose that you used 0.75g of iron. What is the minimum volume of 1.5 M_ CuSO4 solution required
Using eqn. (3-1)? and eqn. (3-2)?

(3-1)
Fe(s) + Cu^2+ (aq) ---- Fe^2 (aq)+ Cu (s)
(3-2)
2Fe(s)+ 3Cu^2+(aq) --- 2Fe^3+ (aq) +3Cu (s)

Answer 1

You don't provide enough information for me to know if you have both equations or just one reacting.
If the first one only, it is
(0.75/55.85) x (1 mol Cu^2+/1 mol Fe) = ?; then M CuSO4 = mols CuSO4/L CuSO4. You know mols and M, solve for L.

If the second one only, it is
(0.75/55.85) x (3 mol Cu^2+/2 mol Fe) =?; then, M CuSO4 = mols CuSO4/L CuSO4. You know mols and M, solve for L CuSO4.

If both then add the volumes CuSO4 together.
Also, if you can provide more information concerning the lab experiment, I may be able to distinguish betwen the two options.