CaCO3 ==> CaO + CO2
mols CO2 liberated = grams/molar mass = estimated 1.32/44 = about 0.03
grams CaCO3 initially present = mols x molar mass = 0.03*100 = about 3 g.
grams CaCO3 + CaO mixture initially = 5.0
grams CaO initially = 5.0-3 = about 2
% CaO = (2/5)*100 = ?
Post your work if you get stuck.
5.0g of a mixture of caco3 and caO liberated 1.32g of carbon(4) oxide on strong heating .
what is the percentage of Cao in mixture?(c=12,O=16,Ca=40)?
2 answers
So do you mean your answer is 40%
1 answer