Lakes that have been acidified by acid rain can be neutralized by the addition of limestone (CaCO3). 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?
Volume of lake is 5.2 × 109-L and Amount of H2SO4 present is = 5.0× 10−3 g
We have to find out the amount of CaCO3 in kg required to neutralize a 5.2 × 109-L lake
5.0×109 L × 5.2×10-3 g/L = 2.6×107 g H2SO4
2.6×107 g H2SO4 / 98.1 g/ mole = 2.65×105 mole H2SO4