One way to raise the pH of an acidic lake (and restore aquatic life) is to add limestone (CaCO3).
How many grams of CaCO3 are needed to neutralize 1.2kL of the acidic water from the lake if the acid is sulfuric acid?
H2SO4(aq)+CaCO3(s)→CO2(g)+H2O(l)+CaSO4(aq)
pH = ? or (H2SO4) = ?
pH=6.5
pH = 6.5 = -log(H^+). Solve for (H^+).
Convert to M H2SO4.
Then mols acid = (H2SO4) x L = ?
Using the coefficients in the balanced equation, convert mols H2SO4 to mols CaCO3, then convert to grams CaCO3.
Well, well, well, looks like we have a chemistry question here! Don't worry, I won't be using any clown chemistry for this one.
To figure out how many grams of CaCO3 are needed to neutralize the acidic water, we need to use stoichiometry. The balanced equation tells us that 1 mole of H2SO4 reacts with 1 mole of CaCO3.
First, we need to find the number of moles of H2SO4 in 1.2 kL of the acidic water. Since 1 kL is equal to 1000 liters, 1.2 kL is equal to 1200 liters.
The concentration of H2SO4 is not given, so we cannot directly calculate the number of moles. However, we can assume that it is a dilute solution and use an approximation.
Let's assume a concentration of 1 mol/L for H2SO4. Therefore, we have 1 mole of H2SO4 per liter. So in 1200 liters, we have 1200 moles of H2SO4.
Since the stoichiometry of the balanced equation tells us that 1 mole of H2SO4 reacts with 1 mole of CaCO3, we need an equal number of moles of CaCO3. So we also need 1200 moles of CaCO3.
Now, we just need to convert moles of CaCO3 to grams using its molar mass. The molar mass of CaCO3 is 100.09 g/mol.
1200 moles x 100.09 g/mol = 120,108 grams
So, approximately 120,108 grams of CaCO3 are needed to neutralize 1.2 kL of the acidic water.
To determine the number of grams of CaCO3 needed to neutralize the acidic water, we need to use stoichiometry and the balanced chemical equation provided. Let's break down the steps to find the answer:
Step 1: Determine the molarity of the sulfuric acid (H2SO4).
Since the given information doesn't mention the molarity of sulfuric acid, we'll need it to proceed. If you have that information, you can skip this step and move on to the next.
Step 2: Convert kiloliters (kL) to liters (L).
1 kiloliter is equal to 1000 liters. Therefore, 1.2 kL is equal to 1200 L.
Step 3: Calculate the number of moles of sulfuric acid (H2SO4).
To determine the number of moles, we need to use the molarity of the sulfuric acid. This is given as M in units of moles per liter (mol/L). Let's assume the molarity of sulfuric acid is x mol/L.
Number of moles of H2SO4 = Molarity (mol/L) * Volume (L)
= x mol/L * 1200 L
= 1200x moles
Step 4: Determine the stoichiometry between H2SO4 and CaCO3.
The balanced chemical equation shows that each mole of sulfuric acid (H2SO4) reacts with one mole of calcium carbonate (CaCO3). Therefore, the stoichiometric ratio between them is 1:1.
Step 5: Calculate the number of moles of CaCO3 required.
Since the stoichiometric ratio is 1:1, the number of moles of CaCO3 required is also equal to 1200x moles.
Step 6: Convert moles of CaCO3 to grams.
To convert moles to grams, we need to know the molar mass of CaCO3. The molar mass of calcium (Ca) is 40.08 g/mol, carbon (C) is 12.01 g/mol, and oxygen (O) is 16.00 g/mol.
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol * 3)
= 100.09 g/mol
Number of grams of CaCO3 = Number of moles * Molar mass
= 1200x moles * (100.09 g/mol)
This equation gives you the number of grams of CaCO3 needed to neutralize the 1.2 kL of acidic water from the lake. Now, substituting the appropriate value for the molarity of the sulfuric acid (H2SO4) will give you the final answer.