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 5.2 x 10^7 -L lake containing 5.0 x 10^(-3) g of H2SO4 per liter.

This is a stoichiometry problem. The equation is

CaCO3 + H2SO4 ==> CaSO4 + CO2 + H2O
If you need a refresher on stoichiometry, here is a solved example.
http://www.jiskha.com/science/chemistry/stoichiometry.html

yaeh dawg this didnt help but thanks

To calculate the amount of limestone required to neutralize the acid in the lake, we need to determine the number of moles of H2SO4 present in the lake and then the corresponding amount of limestone needed.

First, let's calculate the number of moles of H2SO4 in the lake:

Mass of H2SO4 = Concentration of H2SO4 x Volume of the lake
Mass of H2SO4 = (5.0 x 10^(-3) g/L) x (5.2 x 10^7 L)

Now, let's convert the mass of H2SO4 to moles using the molar mass of H2SO4:

Molar mass of H2SO4 = (2 x Atomic mass of H) + Atomic mass of S + (4 x Atomic mass of O)
Molar mass of H2SO4 = (2 x 1.008 g/mol) + 32.06 g/mol + (4 x 16.00 g/mol)
Molar mass of H2SO4 = 98.09 g/mol

Number of moles of H2SO4 = (Mass of H2SO4) / (Molar mass of H2SO4)
Number of moles of H2SO4 = [(5.0 x 10^(-3) g/L) x (5.2 x 10^7 L)] / 98.09 g/mol

Now that we have the number of moles of H2SO4, we can determine the amount of limestone required to neutralize it.

The balanced chemical equation for the reaction between H2SO4 and limestone is as follows:
CaCO3 (limestone) + H2SO4 → CaSO4 + CO2 + H2O

From the equation, we can see that 1 mole of CaCO3 is required to neutralize 1 mole of H2SO4.

Therefore, the amount of limestone required is equal to the number of moles of H2SO4:

Amount of limestone (in moles) = Number of moles of H2SO4

Finally, we need to convert the moles of limestone to kilograms:

Mass of limestone (in kg) = Amount of limestone (in moles) x Molar mass of CaCO3

The molar mass of CaCO3 is 100.09 g/mol.

Mass of limestone (in kg) = (Number of moles of H2SO4) x (100.09 g/mol) / 1000

Now you can substitute the calculated values into the equation to find the amount of limestone required.

To determine how much limestone is required to neutralize the given lake, we need to consider the reaction between limestone (calcium carbonate, CaCO3) and sulfuric acid (H2SO4). The balanced equation for this reaction is:

CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O

From the equation, we can see that one mole of CaCO3 reacts with one mole of H2SO4 to produce one mole of CaSO4, one mole of CO2, and one mole of H2O.

First, let's calculate the number of moles of sulfuric acid (H2SO4) in the lake:

Number of moles of H2SO4 = (5.0 x 10^(-3) g/L) × (5.2 x 10^7 L) = 5.2 x 10^4 g

Next, we need to convert the mass of sulfuric acid to moles:

1 mole of H2SO4 = 98.09 g

Number of moles of H2SO4 = (5.2 x 10^4 g) / (98.09 g/mol) = 530.08 mol

From the balanced equation, we know that one mole of CaCO3 reacts with one mole of H2SO4. Therefore, the number of moles of CaCO3 required to neutralize the lake's sulfuric acid is also 530.08 mol.

The molar mass of CaCO3 is:

40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol

Now, let's calculate the mass of limestone (CaCO3) required:

Mass of CaCO3 = (530.08 mol) × (100.09 g/mol) = 53074.6072 g

To convert the mass to kilograms:

Mass of CaCO3 = 53074.6072 g = 53.0746 kg (rounded to four decimal places)

Therefore, approximately 53.075 kilograms of limestone would be required to completely neutralize the 5.2 x 10^7 L lake containing 5.0 x 10^(-3) g of H2SO4 per liter.