posted by Cristian on .
a 110 g solid sample that is a mixture of CaCO3 and CaCl2 is reacted with 1.50 L of 1.45M HCl. Of the calcium salts, only calcium carbonate reacts with HCl- the reaction forms CaCl2 and CO2. The excess/unreacted HCl requires 0.850 L of 0.543 M NaOH to titrate it to the endpoint.
Determine the mass and mass percent of calcium carbonate in the original solid sample
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
1. mols HCl added = 1.5L x 1.45M = about 2 mol HCl but you need to do it more accurately. That reacts with all of the CaCO3 but doesn't use all of the HCl.
2. Excess HCl requires how many mols. That's 0.850L x 0.543M NaOH = about .46 mols NaOH.
3. mols HCl used = about 2 - 0.46 = about 1.5.
4. Use the coefficients in the balanced equation to convert to mols CaCO3. That is mols HCl x 1/2.
5. mass CaCO3 = mols x molar mass
6. %CaCO3 = (mass CaCO3/mass sample)*100 = ?