The balanced equation for the neutralisation reaction is as follows 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O stoichiometry of NaOH to H₂SO₄ is 2:1 the number of moles of NaOH reacted - 0.126 mol/L x 0.0173 L = 0.00218 mol if 2 mol of NaOH reacts with 1 mol of H₂SO₄ then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄ molarity is the number of moles of solute in 1 L solution therefore if 25 mL contains - 0.00109 mol then 1000 mL contains - 0.00109 mol / 25 mL x 1000 mL = 0.0436 mol/L therefore molarity of H₂SO₄ is 0.0436 M