Question

In an experiment 3.36g of iron fillings were added to excess copper (II) sulphate. Calculate the mass of copper that was deposited. (CU= 63.5, Fe = 56)

Answer 1

Fe_(s) + CuSO_4(aq) → FeSO_4(aq) + Cu_(s)
1mol         1mol
Moles of iron used = ^3.36/₅₆
                           = 0.06moles 
Mole ratio of reaction
Fe : Cu
1 : 1
Moles of Cu produced is 0.06.
Thus mass of copper deposited
= 0.06 x 63.5
= 3.81g