Asked by Double checking ans
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.
Answers
Answered by
DrBob222
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.
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.
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