Gọi CTHH cần tìm là A2O.
Ta có: \(n_{Al}=\dfrac{1,08}{27}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=0,1.1=0,1\left(mol\right)\)
TH1: H2SO4 dư.
PT: \(A_2O+H_2SO_4\rightarrow A_2SO_4+H_2O\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Theo PT: \(n_{H_2SO_4}=n_{A_2O}+\dfrac{3}{2}n_{Al}\Rightarrow n_{A_2O}=0,04\left(mol\right)\)
\(\Rightarrow M_{A_2O}=\dfrac{11,28}{0,04}=282\left(g/mol\right)\)
\(\Rightarrow M_A=133\left(g/mol\right)\)
→ A là Cs.
Vậy: CTHH cần tìm là Cs2O.
TH2: H2SO4 hết.
PT: \(A_2O+H_2SO_4\rightarrow A_2SO_4+H_2O\)
\(A_2O+H_2O\rightarrow2AOH\)
\(2Al+2AOH+2H_2O\rightarrow2AAlO_2+3H_2\)
Theo PT: \(n_{A_2O}=n_{H_2SO_4}+\dfrac{1}{2}n_{AOH}=n_{H_2SO_4}+\dfrac{1}{2}n_{Al}=0,12\left(mol\right)\)
\(\Rightarrow M_{A_2O}=\dfrac{11,28}{0,12}=94\left(g/mol\right)\)
\(\Rightarrow M_A=39\left(g/mol\right)\)
→ A là K.
Vậy: CTHH cần tìm là K2O.
