CaO+H2O--->Ca(OH)2
0,052------0,52--------0,52
Ta có
m\(_{Ba\left(OH\right)2}=\frac{600.14,8}{100}=88,8\left(g\right)\)
n\(_{Ba\left(OH\right)2}=\frac{88,8}{171}=0,52\left(mol\right)\)
Theo pthh
m\(_{CaO}=a=0,52.56=29,12\left(g\right)\)
m\(_{H2O}=600-29,12=570,88\left(g\right)\)