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.

Answer 1

Two equations in two unknowns; solve simultaneously.
Let X = mass CaCO3
Let Y = mass MgCO3
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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.

Answer 2

Thank you. But could you explain how you got that equation for solving for x and y? I need to prove my answer.

Answer 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.