\(n_{Ca}=\dfrac{11,2}{40}=0,28\left(mol\right)\)
\(n_{H_2O}=\dfrac{18}{18}=1\left(mol\right)\)
PTHH: Ca + 2H2O --> Ca(OH)2 + H2
Xét tỉ lệ: \(\dfrac{0,28}{1}< \dfrac{1}{2}\) => Ca hết, H2O dư
PTHH: Ca + 2H2O --> Ca(OH)2 + H2
0,28------------->0,28-->0,28
=> VH2 = 0,28.22,4= 6,272 (l)
mCa(OH)2 = 0,28.74 = 20,72 (g)