If 638.44g CuSO4 reacts with 240.0g NaOH, which is the limiting reactant?

CuSO₄ + 2NaOH → Cu (OH) ₂ + Na₂SO₄

M (CuSO₄) = 159.61 g/mol

m (CuSO₄) = 638.44 g

n (CuSO₄) = m (CuSO₄) / M (CuSO₄)

n (CuSO₄) = 638.44/159.61 = 4.00 mol

M (NaOH) = 40.00 g/mol

m (NaOH) = 240.0 g

n (NaOH) = m (NaOH) / M (NaOH)

n (NaOH) = 240.0/40.00 = 6.00 mol

on the reaction equation

CuSO₄ : NaOH = 1 : 2

in practice

CuSO₄ : NaOH = 4 : 6 = 1 : 1.5 ⇒ NaOH is the limiting reactant