3.
\(CT:A_2O_3\)
\(n_{H_2SO_4}=3a\left(mol\right)\)
\(A_2O_3+3H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2O\)
\(a.........3a......................3a\)
Bảo toàn khối lượng :
\(17+3a\cdot98=57+3a\cdot18\)
\(\Rightarrow a=\dfrac{1}{6}\)
\(M_{oxit}=\dfrac{17}{\dfrac{1}{6}}=102\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow2A+16\cdot3=102\)
\(\Rightarrow A=27\)
\(A:Al\)
\(CT:Al_2O_3\)
\(m_{H_2SO_4}=\dfrac{1}{6}\cdot3\cdot98=49\left(g\right)\)
4.
Gọi: hóa trị M là : n
\(n_{HCl}=0.8\cdot2.5=2\left(mol\right)\)
\(2M+2nHCl\rightarrow2MCl_n+nH_2\)
\(\dfrac{2}{n}.........2.......\dfrac{2}{n}\)
\(M_M=\dfrac{18}{\dfrac{2}{n}}=9n\)
Với : n = 3 \(\Rightarrow M=27\)
\(M:Al\)
\(C_{M_{AlCl_3}}=\dfrac{\dfrac{2}{3}}{0.8}=0.833\left(M\right)\)
