How much limestone (CaCO3) in kilograms would be
required to completely neutralize a lake with a volume of 5.2 x 10^9L . The lake contains 5.0 x 10^-3 moles of H2SO4 per liter?
Multiply the volume of the lake by the concentration, that will give you the moles of sulfuric acid. It will take a similar number of moles of limestone.
To completely neutralize the lake, we need to calculate the amount of limestone (CaCO3) required to react with the sulfuric acid (H2SO4) present in the lake.
First, we need to determine the number of moles of H2SO4 in the lake.
Given:
- Volume of the lake = 5.2 x 10^9 L
- Moles of H2SO4 per liter = 5.0 x 10^-3 moles/L
To find the moles of H2SO4 in the lake, we can multiply the volume of the lake by the moles of H2SO4 per liter:
Moles of H2SO4 = (Volume of the lake) x (Moles of H2SO4 per liter)
= (5.2 x 10^9 L) x (5.0 x 10^-3 moles/L)
= 2.6 x 10^7 moles
Now, let's determine the stoichiometry of the reaction between H2SO4 and CaCO3. The balanced equation is:
H2SO4 + CaCO3 -> CaSO4 + H2O + CO2
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 1 mole of CaCO3.
Therefore, the moles of CaCO3 required to neutralize the lake would be the same as the moles of H2SO4.
Moles of CaCO3 = Moles of H2SO4 = 2.6 x 10^7 moles
To convert moles into kilograms, we need to use the molar mass of CaCO3.
Molar mass of CaCO3 = (1 x atomic mass of Ca) + (1 x atomic mass of C) + (3 x atomic mass of O)
= (1 x 40.08 g/mol) + (1 x 12.01 g/mol) + (3 x 16.00 g/mol)
= 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
= 100.09 g/mol
Now, we can calculate the mass of limestone required:
Mass of limestone = Moles of CaCO3 x Molar mass of CaCO3
= 2.6 x 10^7 moles x 100.09 g/mol
= 2.6 x 10^9 g
To convert grams into kilograms, divide the result by 1000:
Mass of limestone in kilograms = 2.6 x 10^9 g / 1000
= 2.6 x 10^6 kg
Therefore, approximately 2.6 x 10^6 kilograms of limestone (CaCO3) would be required to completely neutralize the lake.
To answer this question, we need to determine the number of moles of H2SO4 in the entire lake and then calculate how much limestone (CaCO3) is required to neutralize it.
Step 1: Calculate the number of moles of H2SO4 in the lake.
Given:
Volume of the lake (V) = 5.2 x 10^9 L
Moles of H2SO4 per liter (n) = 5.0 x 10^-3 mol/L
To calculate the number of moles of H2SO4 in the lake, we can use the formula:
Number of moles (N) = Volume (V) x Moles per liter (n)
N = 5.2 x 10^9 L x 5.0 x 10^-3 mol/L
Step 2: Calculate the mass of CaCO3 required to neutralize the moles of H2SO4.
The balanced chemical equation for the neutralization reaction between H2SO4 and CaCO3 is:
H2SO4 + CaCO3 -> CaSO4 + CO2 + H2O
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 1 mole of CaCO3.
Therefore, the number of moles of CaCO3 required to neutralize the moles of H2SO4 is equal to N.
Step 3: Convert moles of CaCO3 to kilograms.
To convert moles to grams, we can use the molar mass of CaCO3.
The molar mass of CaCO3 is:
- Molar mass of Ca = 40.08 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
Summing up the molar masses:
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (16.00 g/mol x 3) = 100.09 g/mol
To convert moles to kilograms, multiply the number of moles by the molar mass and divide by 1000 (to convert grams to kilograms):
Mass (in kg) = N x Molar mass (in g/mol) / 1000
Finally, plug in the value of N into the equation to calculate the mass of CaCO3 required to neutralize the lake.