a.b.
\(n_{Ba}=\dfrac{24,66}{127}=0,18mol\)
\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
0,18 0,18 ( mol )
\(V_{H_2}=0,18.22,4=4,032l\)
c.
\(n_{CuO}=\dfrac{15,2}{80}=0,19mol\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,19 > 0,18 ( mol )
0,18 0,18 0,18 ( mol )
\(\left\{{}\begin{matrix}m_{Cu}=0,18.64=11,52g\\m_{CuO}=\left(0,19-0,18\right).80=0,8g\end{matrix}\right.\)