What amount of copper will be deposited when 28g of iron filings react with excess copper sulphate solution

Iron reacts with copper(II) sulfate solution in a single displacement reaction:
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
The equation is already balanced, showing a 1:1 mole ratio between iron (Fe) and copper (Cu).

We'll use the molar mass of iron (Fe ≈ 55.845 g/mol).
Moles of Fe = mass / molar mass
Moles of Fe = (28 g) / (55.845 g/mol)
Moles of Fe ≈ 0.5014 mol

From the balanced equation, 1 mole of Fe reacts to produce 1 mole of Cu.
Therefore, moles of Cu = moles of Fe
Moles of Cu ≈ 0.5014 mol

We'll use the molar mass of copper (Cu ≈ 63.546 g/mol).
Mass of Cu = moles × molar mass
Mass of Cu = (0.5014 mol) × (63.546 g/mol)
Mass of Cu ≈ 31.86 g

Rounding to two significant figures (based on the given 28g), the mass of copper deposited is 32 g.

The amount of copper deposited will be 32 g.