How much limestone in kilograms would be required to completely neutralize a 5.2×109 L lake containing 5.0×10−3 g of H2SO4 per liter?

Limestone is calcium carbonate with chemical formula CaCO₃.
Molar mass of CaCO₃ = (40.1 + 12.0 + 16.0×3) g/mol = 100.1 g/mol
Molar mass of H₂SO₄ = (1.0×2 + 32.1 + 16.0×4) g/mol = 98.1 g/mol

CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
OR: Mole ratio CaCO₃ : H₂SO₄ = 1 : 1

No. of moles of H₂SO₄ = (5.0 × 10⁻³ g/L) × (5.2 × 10⁹ L) / (98.1 g/mol) = 2.65 × 10⁵ mol
No. of moles of CaCO₃ required = 2.65 × 10⁵ mol
Mass of CaCO₃ required = (2.65 × 10⁵ mol) × (100.1 g/mol) = 2.65 × 10⁷ g = 2.65 × 10⁴ kg = 26500 kg