The copper (II) sulfate will be the excess reactant by 6 g.
Start with the balanced chemical equation
Notice that you have a
Whichever reactant will have more moles present, that will be your excess reactant. Determine how many moles of each reactant are present by using their respective molar mass
Since you have more moles of copper (II) sulfate than you have of iron, copper (II) sulfate will be the excess reactant (iron will be the limiting reagent).
The reaction will consume all of the iron and leave you with
This is equivalent to having an excess of
Rounded to one sig fig, the number of sig figs given for the mass of iron, the answer will be