A 1.00 g sample of dolomite containing 23% calcite , 42% magnesite and the rest water, is analyzed by..?
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thermally decomposing the sample at 800K until all the magnesium carbonate has decomposed. (none of the calcite decomposes) In addition to the weight loss due to the production of carbon dioxide, all the water associated with the sample is irreversibly lost.
Calculate the weight of the sample after calcinations.
Note: calcite is calcium carbonate and magnesite is magnesium carbonate
Can anyone help me with this?
calcite = CaCO3
magnesite = MgCO3
It’s been a few years since I studied alchemy, but I think this might be happening:
MgCO3*H2On –> MgO + CO2 + nH2O
CaCO3*H2Om –> CaCO3 + mH2O
Weight of calcite:
1.00g x 23% [ when you see percent, it's by weight or by volume ]
Weight of magnesium oxide:
1.00g x 42% [weight of the magnesite]
x (1 mol MgCO3) / (??? g MgCO3) [ molecular weight of magnesite ]
x (1 mol MgO) / (1 mol MgCO3) [ stoichiometry; see above ]
x (??? g MgO) / (1 mol MgO) [molecular weight of magnesium oxide ]
Add the two weights. Done.
IM has dusted off those chemistry brain cells nicely. When you have calculated the remaining solids you will get 0.43 grams of CaCO3 and MgO.
What concerns me is your use of the plural in “calcinations”. Are you implying that both of the carbonate minerals are decomposed? If so, when you’ve done all the calculations then you will see 0.33 grams of MgO and CaO.
You may know that the word “calcination” refers to the “cooking” of limestone, CaCO3, to form lime, CaO.