Gọi CTHH cần tìm là AO.
PT: \(AO+H_2SO_4\rightarrow ASO_4+H_2O\)
Giả sử: mAO = a (g)
\(\Rightarrow n_{AO}=\dfrac{a}{M_A+16}\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{ASO_4}=n_{AO}=\dfrac{a}{M_A+16}\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=\dfrac{98a}{M_A+16}\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}}{9,8\%}=\dfrac{1000a}{M_A+16}\left(g\right)\)
⇒ m dd sau pư = \(a+\dfrac{1000a}{M_A+16}\) (g)
Có: \(m_{ASO_4}=\dfrac{\left(M_A+96\right).a}{M_A+16}\left(g\right)\)
\(\Rightarrow\dfrac{\dfrac{\left(M_A+96\right).a}{M_A+16}}{a+\dfrac{1000a}{M_A+16}}=0,1154\)
⇒ MA = 24 (g/mol)
→ A là Mg
Vậy: CTHH cần tìm là MgO.
