Astudent decomposed 3.67g of copper (ii) hydroxide into copper (ii) oxide. how many ml of 3m h2so4 is need to react with all the copper (ii) oxide?

In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide. 
Cu(OH)₂ (s) > CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂  - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄ 
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present 
so if 3 mol are present in 1000 ml 
then volume for 0.038 mol = 1000/3 * 0.038 
                                           = 12.67 ml