$a)n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)$

$n_{CuO}=\dfrac{40}{80}=0,5(mol)$

$PTHH:CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Xét tỉ lệ: $n_{CuO}>n_{H_2}\Rightarrow CuO$ dư

$\Rightarrow n_{CuO(dư)}=0,5-0,25=0,25(mol)$

$\Rightarrow m_{CuO(dư)}=0,25.80=20(g)$

$b)$ Chất rắn thu đc gồm: $CuO(dư)$ và $Cu$

Theo PT: $n_{Cu}=n_{H_2}=0,25(mol)$

$\Rightarrow m_{Cu}=0,25.64=16(g)$

$\Rightarrow m_{rắn}=16+20=36(g)$

$\Rightarrow \%m_{Cu}=\dfrac{16}{36}.100\%=44,44\%$

$\Rightarrow \%m_{CuO}=100-44,44=55,56\%$

$c)n_{CuO(p/ứ)}=n_{H_2}=0,25(mol)$

$H\%=\dfrac{m_{CuO(p/ứ)}}{n_{CuO(bđ)}}.100\%=\dfrac{0,25}{0,5}.100\%=50\%$