\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Ca + 2H2O -------> Ca(OH)2 + H2
0,1 0,1 0,1
CaO + H2O -------> Ca(OH)2
\(\%m_{Ca}=\dfrac{0,1.40.100\%}{9,6}=41,67\%\)
\(\%m_{CaO}=100\%-41,67\%=58,33\%\)
\(n_{Ca\left(OH\right)_2}=n_{Ca}+n_{CaO}=0,1+\dfrac{9,6-0,1.40}{56}=0,2\left(mol\right)\)
\(m_{Ca\left(OH\right)_2}=0,2.84=16,8\left(g\right)\)