How much limestone (CaCO3) in kilograms would be
required to completely neutralize a lake with a volume of 5.2 x 10^7 -L lake containing 5.0 x 10-3g of H2SO4 per liter?
To find out how much limestone is required to completely neutralize the lake, we can follow these steps:
Step 1: Calculate the total mass of H2SO4 in the lake.
Given:
Volume of the lake = 5.2 x 10^7 L
Concentration of H2SO4 = 5.0 x 10^-3 g/L
To find the total mass of H2SO4, we can multiply the concentration by the volume:
Total mass of H2SO4 = Concentration of H2SO4 x Volume of the lake
= (5.0 x 10^-3 g/L) x (5.2 x 10^7 L)
= 2.6 x 10^5 g
Step 2: Convert the mass of H2SO4 to moles.
To convert grams to moles, we need to divide the mass by the molar mass of H2SO4. The molar mass of H2SO4 is the sum of the atomic masses of hydrogen (H), sulfur (S), and oxygen (O).
Molar mass of H2SO4 = (2 x Molar mass of H) + Molar mass of S + (4 x Molar mass of O)
= (2 x 1.01 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)
= 98.09 g/mol
Number of moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4
= 2.6 x 10^5 g / 98.09 g/mol
= 2.65 x 10^3 moles
Step 3: Determine the stoichiometry of the neutralization reaction.
The balanced chemical equation for the neutralization of H2SO4 and CaCO3 is given as:
H2SO4 + CaCO3 -> CaSO4 + H2O + CO2
From the equation, we can see that 1 mole of H2SO4 reacts with 1 mole of CaCO3.
Step 4: Calculate the amount of CaCO3 required.
Since 1 mole of H2SO4 reacts with 1 mole of CaCO3, the number of moles of CaCO3 required will be the same as the number of moles of H2SO4.
Number of moles of CaCO3 required = Number of moles of H2SO4
= 2.65 x 10^3 moles
Step 5: Convert moles of CaCO3 to mass.
To convert moles to mass, we need to multiply the number of moles by the molar mass of CaCO3. The molar mass of CaCO3 is the sum of the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms.
Molar mass of CaCO3 = Molar mass of Ca + Molar mass of C + (3 x Molar mass of O)
= 40.08 g/mol + 12.01 g/mol + (3 x 16.00 g/mol)
= 100.09 g/mol
Mass of CaCO3 required = Number of moles of CaCO3 x Molar mass of CaCO3
= 2.65 x 10^3 moles x 100.09 g/mol
= 2.65 x 10^5 g
Therefore, approximately 2.65 x 10^5 grams or 265 kilograms of limestone (CaCO3) would be required to completely neutralize the lake.