a. PTHH (1) Ca+H2O->CaO+H2\(\uparrow\)
mol 1 1 1 1
(2) CaO+H2O->Ca(OH)2
mol 1 1 1
n\(_{H^{ }_2}\)=\(\frac{3,36}{22,4}\)=0,15 mol
Theo phương trình ta có: n\(_{Ca}\)=n\(_{H_2}\)=0,15 mol
=>m\(_{Ca}\)=0,15x40=6 g
=>m\(_{CaO}\)=17,2-6=11,2 g
b. %m\(_{Ca}\)=\(\frac{m_{Ca}}{m_{hh}}\)x100%=\(\frac{6}{17,2}\)x100%=34,88%
=>%m\(_{CaO}\)=100%-34,88%=65,12%