Posted by Double checking ans on .
A CaCO3 mixture weighing 2.75 grams was treated with 9.75 mL of 3.00 M HCl to liberate CO2. After the CO2 was liberated, the mass of the mixture decreased to 2.621 grams. Determine the mass of CaCO3 present in the mixture.
Chem 151 -
Loss in mass = mass CO2 = 2.621g.
mols CO2 = grams CO2/molar mass CO2.
mols CaCO3 = mols CO2 (look at the balanced equation).
CaCO3 ==> CaO + CO2.
g CaCO3 = mols CaCO3 x molar mass CaCO3.