Question

2.5 g of iron powder are added to 100 mL of a 0.75 mol/L copper (ll) sulfate solution

1- determine the composition of the solution in molar concentration when the reaction ceases?

2_ what is the mass of copper produced

Answer 1

Fe(s) + CuSO4(aq) ==> FeSO4(aq) + Cu(s)

mols CuSO4 initially = M x L = 0.75 M x 0.100 L = 0.075

mols Fe initially = g/atomic mass = 2.5/55.85 = 0.045

This is a limiting reagent problem in which mols Fe are depleted completed and there is some CuSO4 remaining after the reaction ceases.

So how much Cu is formed. That is

0.045 mols Fe x (1 mol Cu/1 mol Fe) = 0.045 mols Cu produced.

grams Cu = 0.045 mols Cu x 63.54 g Cu/mol = 2.8 g Cu.