\(m_{d^2\text{ }sau\text{ }pứ}=m_{M_xO_y}+m_{d^2\text{ }HCl}=2,08+100=102,08\left(g\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{102,08\cdot0,7866}{100}=0,8\left(g\right)\)
\(\Sigma m_{HCl}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{3,65\cdot100}{100}=3,65\left(g\right)\\ \Rightarrow m_{HCl\left(pứ\right)}=3,65-0,8=2,85\left(g\right)\\ \Rightarrow n_{HCl\left(pứ\right)}=\dfrac{m}{M}=\dfrac{2,85}{36,5}=0,078\left(mol\right)\)
\(pthh:M_xO_y+2yHCl\rightarrow xMCl_{\dfrac{2y}{x}}+yH_2O\left(1\right)\)
Theo \(pthh\left(1\right):n_{M_xO_y}=\dfrac{1}{2y}n_{HCl}=\dfrac{0,078}{2y}=\dfrac{0,039}{y}\)
\(\Rightarrow M_{M_xO_y}=\dfrac{m}{n}=\dfrac{2,08}{\dfrac{0,039}{y}}=\dfrac{160}{3}y\\ \Rightarrow x\cdot M_M+16y=\dfrac{160}{3}y\\ \Rightarrow M_M=\dfrac{56}{3}\cdot\dfrac{2y}{x}\)
Vì M là kim loại nên ta có bảng biện luận:
| \(\dfrac{2y}{x}\) | \(1\) | \(2\) | \(3\) | \(4\) | \(\dfrac{8}{3}\) |
| \(M_M\) | \(\dfrac{56}{3}\left(g\right)\rightarrow loại\) | \(\dfrac{112}{3}\left(g\right)\rightarrow loại\) | \(56\left(g\right)\rightarrow Fe\) | \(\dfrac{224}{3}\left(g\right)\rightarrow loại\) | \(\dfrac{448}{9}\left(g\right)\rightarrow loại\) |
\(\Rightarrow\) M là kim loại \(Fe\left(sắt\right)\)
\(\Rightarrow\dfrac{2y}{x}=3\\ \Rightarrow\dfrac{y}{x}=\dfrac{3}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
\(\Rightarrow M_xO_y=Fe_2O_3\)
Vậy CTHH cần lập là \(Fe_2O_3\)
