a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH:
Ca + 2H2O ---> Ca(OH)2 + H2
0,1<-----------------------------0,1
\(m_{CaO}=6-0,1.40=2\left(g\right)\\ m_{Ca}=0,1.40=4\left(g\right)\\ \left\{{}\begin{matrix}\%m_{Ca}=\dfrac{4}{6}.100\%=66,67\%\\\%m_{CaO}=100\%-66,67\%=33,33\%\end{matrix}\right.\)
