The idea here is that you add an excess of HCl to the sea shells and that completely reacts with ALL of the calcium carbonate. The excess HCl is titrated with NaOH.
CaCO3 + 2HCl -> CaCl2 + 2H2O
mols HCl added initially = M x L = 0.025 x 5 = 0.125.
How much HCl was in excess? That's 0.0436 x 2.2 = 0.0959
Here's what we have.
0.125 mols HCl initially
-0.0959 mols xs HCl
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0.0291 mols HCl used by the CaCO3.
Convert mols HCl to mols CaCO3.
0.0291mols HCl x (1 mol CaCO3/2 mols HCl) = 0.0291/2 = ?mols CaCO3
Convert to grams. g = mols CaCO3 x molar mass CaCO3.
Then %CaCO3 = (mass CaCO3/mass sample)*100 - ?
A Chemist wanted to determine the percentage of calcium carbonate in a sea shell. He collected a number of sea shells which he crushed into a fine powder.
To 2.65 grams of sea shell powder he added 25.0mL of 5.0M Hydrochloric acid.
When all the fizzing ceased, he filtered the remaining mixture into a conical flask and carefully washed the residue.
Finally he titrated the filtrate with 2.20M Sodium Hydroxide and 46.3mL of base was required.
Calculate the percentage of calcium carbonate in the sea shells.
1 answer