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 x 109 L lake containing 5.0 x 10-3 g of H2SO4 per liter?

Same process as the stoichiometry problem after you determine how much cid is in the lake. That is 5.2 x 10^9 L x 5.0E-3 g/L = ??

To calculate the amount of limestone (CaCo3) needed to neutralize the lake, we need to determine the amount of acid (H2SO4) that needs to be neutralized.

1. Determine the total amount of acid in the lake:
The concentration of H2SO4 is given as 5.0 x 10^-3 g/L.
Multiply the concentration by the volume of the lake to find the total amount:
Total amount of H2SO4 = (5.0 x 10^-3 g/L) x (5.2 x 10^9 L)

Note: You need to ensure the units match. In this case, we multiply grams per liter by liters to get grams.

2. Convert the mass of H2SO4 to moles:
- Determine the molar mass of H2SO4:
H: 1.01 g/mol x 2 = 2.02 g/mol
S: 32.07 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total molar mass = 2.02 g/mol + 32.07 g/mol + 64.00 g/mol = 98.09 g/mol

- Convert the mass of H2SO4 to moles using the molar mass:
Moles = mass (g) / molar mass (g/mol)

3. Determine the stoichiometric ratio between H2SO4 and CaCO3:
From the balanced equation:
H2SO4 + CaCO3 → CaSO4 + H2O + CO2
The ratio of H2SO4 to CaCO3 is 1:1. One mole of H2SO4 reacts with one mole of CaCO3.

4. Determine the amount of CaCO3 required:
The moles of CaCO3 needed will be equal to the moles of H2SO4:
Moles of CaCO3 = Moles of H2SO4

5. Convert the moles of CaCO3 to mass:
Mass (g) = Moles x molar mass (g/mol)

6. Finally, convert the mass to kilograms:
Mass (kg) = Mass (g) / 1000

Now let's gather the numbers and perform the calculations:

1. Total amount of H2SO4 = (5.0 x 10^-3 g/L) x (5.2 x 10^9 L)
= 26,000,000 g

2. Moles of H2SO4 = 26,000,000 g / 98.09 g/mol
≈ 265,217 moles

3. Moles of CaCO3 = 265,217 moles

4. Mass of CaCO3 = 265,217 moles x 100.09 g/mol (molar mass of CaCO3)
≈ 26,536,956 g

5. Mass of CaCO3 in kilograms = 26,536,956 g / 1000 kg/g
≈ 26,537 kg

Therefore, approximately 26,537 kilograms of limestone (CaCO3) would be required to completely neutralize the 5.2 x 10^9 L lake containing 5.0 x 10^-3 g of H2SO4 per liter.