The active ingredient of the antacid is CaCO3 and so Sally chose titration with HCl as a method for analysis. What is the mass (in g) of the CaCO3 in the sample of antacid, if it is titrated by 21.2 mL of HCl solution with concentration of 0.1048 mol/L to reach the endpoint?
moles acid used = M x L.
Moles CaCO3 is 1/2 that because
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
moles CaCO3 = grams/molar mass.
You have moles and molar mass. Solve for grams.
To find the mass of CaCO3 in the sample of antacid, we'll use the concept of stoichiometry.
First, we need to find the number of moles of HCl used in the titration. To do this, we'll use the formula:
moles = concentration (mol/L) x volume (L)
Given that the concentration of HCl is 0.1048 mol/L and the volume used is 21.2 mL (or 0.0212 L), we can calculate the moles of HCl used:
moles HCl = 0.1048 mol/L x 0.0212 L = 0.002215 moles
Since the balanced chemical equation for the reaction between CaCO3 and HCl is:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
We can see that 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, the number of moles of CaCO3 can be calculated using the mole ratio:
moles CaCO3 = (1/2) x moles HCl = (1/2) x 0.002215 moles = 0.001108 moles
Now, to find the mass of CaCO3, we'll use its molar mass. The molar mass of CaCO3 can be calculated by adding the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms:
Molar mass of CaCO3 = (atomic mass of Ca) + (atomic mass of C) + (3 x atomic mass of O)
= 40.08 g/mol + 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
Finally, we can calculate the mass of CaCO3 by multiplying the number of moles by the molar mass:
mass CaCO3 = moles CaCO3 x molar mass CaCO3
= 0.001108 moles x 100.09 g/mol
≈ 0.1101 g
Therefore, the mass of CaCO3 in the sample of antacid is approximately 0.1101 grams.