`a,`
`n_{Mg}=\frac{3,6}{24}=0,15(mol)`
`Mg+2HCl->MgCl_2+H_2`
Theo phương trinh
`n_{H_2}=n_{MgCl_2}=0,15(mol)`
`->V_{H_2}=0,15.22,4=3,36(l)`
`b,m_{MgCl_2}=0,15.85=14,25(g)`
`c,`
$CuO+H_2\xrightarrow{t^o}Cu+H_2O$
Theo phương trình
`n_{Cu}=n_{H_2}=0,15(mol)`
`->m_{Cu}=0,15.64=9,6(g)`