What is the excess reactant?

Question

At the beginning of the reaction, it is measured that there are 3.11 g of iron and 4.13 g of copper(II) sulfate.

​​Fe+CuSO4→Cu+FeSO4

What is the excess reactant? 

Answer 1

​​Fe + CuSO₄→Cu + FeSO₄

First, find out how many moles of each reactant are present:

moles Fe = 3.11 g Fe x 1 mol Fe / 55.85 g Fe = 0.0557 moles Fe

moles CuSO₄ = 4.13 g CuSO₄ x 1 mol CuSO₄ / 159.6 g CuSO₄ = 0.0259 moles CuSO₄

Next, look at the balanced equation and note that the reactants react in a mole ratio of 1 : 1, so therefore the limiting reactant (the one that runs out first) is CuSO4. This makes the excess reactant = Fe