Ask questions and get helpful answers.

A mixture of CaCO3 and MgCO3 weighing 7.85 g was reacted with excess HCL. The reactions are CaCO3+2HCL=CaCl2+H2O+CO2 and MgCO3+2HCl=MgCl2+H2O+CO2. The sample reacted completely to produce 1.94 L CO2 at 25 degrees C and 785 torr. Calculate the percentage of CaCO3 and MgCO3 in the original sample.

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
3 answers
  1. Two equations in two unknowns; solve simultaneously.
    Let X = mass CaCO3
    Let Y = mass MgCO3
    ======================
    X + Y = 7.85 grams.
    X(moles mass CO2/molar mass CaCO3) + Y(molar mass CO2/molar mass MgCO3) = grams CO2. [Note: YOu will need to use PV = nRT to calculate moles CO2, then convert that to grams CO2.]
    Solve for X and Y then apply
    %CaCO3 = (mass CaCO3/mass sample)*100 = ??
    %MgCO3 = (mass MgCO3/mass sample)*100 = ??
    Post your work if you get stuck.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. Thank you. But could you explain how you got that equation for solving for x and y? I need to prove my answer.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. .819 mol CO2= 30604 g CO2
    X= 3.604 g CO2/100.09 g/molCaCO3= .036 mol CaCO3
    Y=3.604 g CO2/84.32 g/mol MgCO3= .043 mol MgCO3

    I'm stuck. I don't think I did it right.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.