Find the volume of 0.130 M sulfuric acid necessary to react completely with 76.1 g sodium hydroxide.?

Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Number of moles of NaOH reacted = (76.1 g) / (40.0 g/mol) = 1.9025 mol

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Mole ratio NaOH : H₂SO₄ = 2 : 1
Number of H₂SO₄ needed = (1.9025 mol) × (1/2) = 0.95125 mol
Volume of H₂SO₄ needed = (0.95125 mol) / (0.130 mol/L) = 7.32 L

OR:
(76.1 g NaOH) × (1 mol NaOH / 40.0 g NaOH) × (1 mol H₂SO₄ / 2 mol NaOH) × (1 L H₂SO₄ / 0.130 mol H₂SO₄)
= 7.32 L