Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
a, PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2 (1)
b. Ta có: \(n_{H_2}=\dfrac{3.0,2}{2}=0,3\left(mol\right)\\
n_{HCl}=\dfrac{6.0,2}{2}=0,6\left(mol\right)\)
=> \(V_{H_2\left(đktc\right)}=0,6.22,4=13,44\left(l\right)\)
c, \(V_{ddHCl}=\dfrac{n_{HCl}}{C_{MddHCl}}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)
d, PTHH: MO + H2 -to-> M + H2O (2)
Ta có: \(n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,3\left(mol\right)\\ n_{MO\left(2\right)}=n_{H_2\left(2\right)}=0,3\left(mol\right)\\ =>M_{MO}=\dfrac{24}{0,3}=80\left(\dfrac{g}{mol}\right)->\left(1\right)\\ Mà:M_{MO}=M_M+M_O=M_M+16->\left(2\right)\\ Từ\left(1\right),\left(2\right)=>M_M+16=80\\ =>M_M=80-16=64\left(nhận:Cu\right)\\
\)
=> CTHH của oxit kim loại M đó là CuO.