nHCl = 0,8.1=0,8(mol)
CTHH: RxOy
PTHH: \(R_xO_y+2yHCl->xRCl_{\dfrac{2y}{x}}+yH_2O\)
_______\(\dfrac{0,4}{y}\)<----0,8____________________(mol)
=> \(M_{R_xO_y}=\dfrac{23,2}{\dfrac{0,4}{y}}=58y\left(g/mol\right)\)
=> x.MR = 42y => \(M_R=21.\dfrac{2y}{x}\)
Xét \(\dfrac{2y}{x}=1\) => MR = 21 (L)
Xét \(\dfrac{2y}{x}=2\) => MR = 42 (L)
Xét \(\dfrac{2y}{x}=3=>M_R=63\left(L\right)\)
Xét \(\dfrac{2y}{x}=\dfrac{8}{3}=>M_R=56\left(Fe\right)\)