a, PTHH: Ca+2H2O--->Ca(OH)2+H2 (1)
CaO+H2O--->Ca(OH)2 (2)
nH2= \(\dfrac{2,24}{22,4}=0,1\) mol
Theo pt(1) nH2=nCa=0,1 mol
=> mCa= 0,1.40= 4 g
mCaO= 17,2-4=13,2 g
=> %Ca= \(\dfrac{4}{17,2}.100\%\approx23,26\%\)
%CaO= 100%-23,26%= 76,74%
b, Theo pt(1) và (2): nCa(OH)2=nCaO= 0,1 mol
=> mCa(OH)2= 0,1.74= 7,4 g