Question

A 5.000 g sample of a mixture which contains MgCO3 and sand (SiO2) was heated for 2.30 hours until no further reaction occurred. It was then cooled and the residue, which contained MgO and unchanged sand, was weighed. The reaction is: MgCO3(s)  MgO(s) + CO2(g). The residue remaining (minus the CO2 which was completely driven off) weighed 3.397 grams. Calculate the percent, by weight, of MgCO3 in the original sample.

Answer 1

MgCO3 + SiO2 + heat ==> MgO + CO2 + SiO2

Loss in mass = 5.000-3.397 = 1.603 g = mass CO2.
mols CO2 = 1.603/44 = ?
mols MgCO3 = the same (from the equation).
mass MgCO3 = mols x molar mass
%MgCO3 = (mass MgCO3/mass sample)*100 = ?

Answer 2

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