the balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of H₂SO₄ moles reacted - 0.500 mol/L x 1.50 L = 0.750 mol
1 mol of H₂SO₄ reacts with 2 mol of NaOH
therefore 0.750 mol of H₂SO₄ reacts with - 2 x 0.750 = 1.50 mol of NaOH
molarity of NaOH solution is 0.750 M
there are 0.750 mol in 1 L of solution
therefore 1.50 mol in - 1.50 mol / 0.750 mol/L = 2 L
2 L of NaOH is required for neutralisation
