A .1360 g sample of a mixture of magnesium and calcium carbonates when treated with HCL evolves 32.25 cm cubed of carbon dioxed measured dry at STP. Calculate the percent of magnesium carbonate in the mixutre.

Answer 1

I am going to show you how to do this problem using chemical factors to convert from one material to another. This isn't taught in classes anymore but it sure is convenient.
Let x = mass MgCO3
then y = mass CaCO3
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X + Y = 0.1360g
X(molar mass CO2/molar mass MgCO3) + Y(molar mass CO2/molar mass CaCO3) = (0.03225*44/22.4)
Solve the two equations simultaneously for X and Y (actually, X is all you need). Then %MgCO3 =(mass MgCO3/mass sample)*100 = ??
[Note: molar mass CO2/molar mass MgCO3 convert X(grams MgCO3) to grams CO2. Similarly, molar mass CO2/molar mass CaCO3 conerts Y(grams CaCO3) to grams CO2. Both of these are chemical factors. On the right hand side, the 32.25 cc converts to 0.03225L and that divided by 22.4 converts to moles CO2. That multiplied by 44 converts to grams CO2 so basically we have the equation 2 just grams CO2 in terms of X + grams CO2 in terms of Y = grams CO2 attained.