Lakes that have been acidified by acid rain can be

neutralized by the addition of limestone
How much limestone in kilograms would be
required to completely neutralize a
lake containing of H2 per liter? 5.0 * 10-3 g SO4
5.2 *

To determine the amount of limestone required to neutralize the lake, we need to calculate the amount of H2SO4 that needs to be neutralized and then convert it to kilograms.

Given:
- The concentration of H2SO4 in the lake is 5.0 * 10^-3 g/l.
- The molar mass of H2SO4 is 98.09 g/mol.
- The balanced chemical equation for neutralization: CaCO3 (limestone) + H2SO4 → CaSO4 + CO2 + H2O.

Step 1: Convert the given concentration of H2SO4 to moles per liter (mol/l).
- 5.0 * 10^-3 g/l = 5.0 * 10^-6 kg/l.
- Since the molar mass of H2SO4 is 98.09 g/mol, we can calculate the concentration in moles per liter: 5.0 * 10^-6 kg/l / 98.09 g/mol = 5.1 * 10^-8 mol/l.

Step 2: Calculate the amount of H2SO4 in the entire lake.
- Multiply the concentration by the volume of the lake in liters. Let's assume the volume of the lake is V liters.
- Amount of H2SO4 = 5.1 * 10^-8 mol/l * V l = (5.1 * 10^-8 V) mol.

Step 3: Determine the stoichiometry between H2SO4 and CaCO3.
From the balanced chemical equation, we can see that 1 mol of H2SO4 reacts with 1 mol of CaCO3. Therefore, the amount of CaCO3 needed is equal to the amount of H2SO4.

Step 4: Convert the amount of CaCO3 to kilograms.
- The molar mass of CaCO3 is 100.09 g/mol.
- Amount of CaCO3 = (5.1 * 10^-8 V) mol.
- Convert this to kilograms: (5.1 * 10^-8 V mol) * (100.09 g/mol) * (1 kg/1000 g) = (5.1 * 10^-10 V) kg.

Therefore, the amount of limestone (CaCO3) required to neutralize the lake is approximately (5.1 * 10^-10 V) kilograms.