Question

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

Answer 1

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

Answer 2

NaOH is the limiting reactant

Step-by-step explanation:

Considering the chemiacl reaction :

CuSO₄ + 2NaOH → Cu(OH)₂ + Na₂SO₄

Molar Mass CuSO₄ = 159 g/mol

Molar mass NaOH = 40 g/mol

Due to stoichiometry :

159 g CuSO₄ = 80 g NaOH

Then: 638.44 g CuSO₄ should react with 321.23 g NaOH, but we have 240.o g of NaOH, so this last is the limiting reactant