The idea here is that you have a ppt of BaSO4 + CaCO3. On heating the CaCO3 loses CO2 but CaO remains. How much remains? That will be 2.40 mg (see below).
If you have 4.3 mg CaCO3, it will lose CO2 in heating. How much will it lose?
CaCO3 ==> CaO + CO2
4.3mg.....2.40..1.89
Converting CaCO3 to CaO and CO2 this way is
4.3 mg CaCO3 x (molar mass CaO/molar mass CaCO3) = about 2.40 mg CaO.
4.3 mg CaCO3 x (molar mass CO2/molar mass CaCO3) = about 1.89 mg CO2.
If the heated sample had a mass of 277 mg BaSO4 + CaO, the amount BaSO4 must be 277-2.40 = about 275 mg.
part b. mg SO4 = about 275 x (molar mass SO4/molar mass BaSO4) = ? AND
%SO4 = (?/800)*100 = x?
part a.
277 mg = mass BaSO4 + CaO
add in 1.89 for mass CO2 lost when heated.
mass sample before heating was 278.9 mg.
Then convert to SO4 as
mg SO4 = 278.9 x (molar mass SO4/molar mass BaSO4) = ?
apparent %SO4 = (?/800)*100 = x?
I haven't paid attention to significant figures. You should.
For part c, I don't know what Ea stands for.
An 800-mg sample containing sulfate was treated with slight excess of barium chloride, yielding a precipitate that contained 4.3 mg of co precipitated CaCO3. after ignition and cooling, the precipitate weighed 277.0 mg. Calculate the a.) apparent % SO4 b.) true % SO4 and c.) Ea in %SO4
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