A 0.5215 gram sample of CaCO3 is dissolved in 12M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask.
How many moles of CaCO3 are used?
What is the molarity of the Ca2+ in the 250 mL of solution?
How many moles of Ca2+ ae in a 25.0 mL aliquot of the unknown solution from question 2?
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To find the number of moles of CaCO3 used, we need to use the given mass and the molar mass of CaCO3.
1. Find the molar mass of CaCO3:
CaCO3 = (40.1 g/mol for Ca) + (12.0 g/mol for C) + (3 * 16.0 g/mol for O) = 100.1 g/mol
2. Use the given mass of CaCO3 to find the number of moles:
moles = mass / molar mass
moles = 0.5215 g / 100.1 g/mol
moles ≈ 0.005211 mol
So, 0.005211 moles of CaCO3 are used.
Next, let's find the molarity of the Ca2+ ions in the 250.0 mL of solution.
To do that, we need to understand the reaction between CaCO3 and HCl.
CaCO3 reacts with HCl to form Ca2+ ions (Ca^2+) and CO2 gas (CO2).
The balanced chemical equation for the reaction is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
From the equation, we can see that 1 mole of CaCO3 produces 1 mole of Ca^2+.
3. Calculate the molarity of Ca2+ ions:
Molarity (M) = moles of solute / volume of solution in liters
We are given the volume of the solution in milliliters, but we need to convert it to liters:
volume of solution = 250.0 mL = 250.0 mL * (1 L / 1000 mL) = 0.250 L
Molarity (Ca^2+) = moles of Ca^2+ / volume of solution in liters
Molarity (Ca^2+) = 0.005211 mol / 0.250 L
Molarity (Ca^2+) ≈ 0.020844 M
Therefore, the molarity of Ca2+ in the 250 mL of solution is approximately 0.020844 M.
Lastly, let's find the number of moles of Ca2+ ions in a 25.0 mL aliquot of the unknown solution from the previous question.
4. Calculate the moles of Ca2+ in the 25.0 mL aliquot:
moles = Molarity (Ca^2+) * volume of aliquot in liters
volume of aliquot = 25.0 mL = 25.0 mL * (1 L / 1000 mL) = 0.025 L
moles = 0.020844 mol/L * 0.025 L
moles ≈ 0.0005211 mol
Therefore, there are approximately 0.0005211 moles of Ca2+ ions in the 25.0 mL aliquot of the solution.