CaCO3 \(\underrightarrow{t^o}\)CaO + CO2
mGiảm=50.22%=11(g)
nCO2=\(\dfrac{11}{44}=0,25\left(mol\right)\)
Theo PTHH ta cso:
nCaCO3=nCO2=0,25(mol)
mCaCO3=100.0,25=25(g)
b;
mCaCO3 sau PƯ=50-25=25(g)
mCaO=25-11=14(g)
mrắn=25+14=39(g)
%mCaCO3=\(\dfrac{25}{39}.100\%=64,1\%\)
%mCaO=100-64,1=35,9%