Monday
March 27, 2017

Post a New Question

Posted by 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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question