nH2 = 3.36 / 22.4 = 0.15 (mol)
a) PTHH : CaO + H2O --> Ca(OH)2
                 Ca + H2O --> CaO + H2
                0.15     0.15      0.15     0.15    (mol)
mCa = 0.15 * 40 = 6 (g)
mCaO = 17.2 - 6 = 11.2 (g)
b) %Ca = 6 / 17.2 * 100 = 35%
    %CaO = 100% - 35% = 65%

a. PTHH (1) Ca+H2O->CaO+H2↑
mol 1 1 1 1
(2) CaO+H2O->Ca(OH)2
mol 1 1 1
nH2H2=3,36/22,4=0,15 mol
Theo phương trình ta có: nCa=nH2=0,15 mol
=>mCa=0,15x40=6 g
=>mCaOCaO=17,2-6=11,2 g
b. %mCaCa=mCa/mhh x100%=6/17,2 x100%=34,88%
=>%mCaO=100%-34,88%=65,12%