Đáp án:

$F{e_2}{(S{O_4})_3}.9{H_2}O$

Giải thích các bước giải:

 PTHH:

$2MS + \left( {2 + \dfrac{n}{2}} \right){O_2}\xrightarrow{{{t^o}}}{M_2}{O_n} + 2S{O_2}$

$\begin{gathered}   {M_2}{O_n} + n{H_2}S{O_4} \to {M_2}{(S{O_4})_n} + n{H_2}O \hfill \\  \end{gathered} $

Gọi a là số mol $MS$

${n_{{M_2}{O_n}}} = {n_{{M_2}{{(S{O_4})}_n}}} = \dfrac{a}{2};{n_{{H_2}S{O_4}}} = \dfrac{{an}}{2}$

$\begin{gathered}   {m_{{M_2}{O_n}}} = \dfrac{a}{2}(2M + 16n);{m_{{M_2}{{(S{O_4})}_n}}} = \dfrac{a}{2}(2M + 96n) \hfill \\   {m_{dd{H_2}S{O_4}}} = \dfrac{{98.\dfrac{{an}}{2}}}{{36,75\% }} = 133,33an \hfill \\  \end{gathered} $

Bảo toàn khối lượng:  ${m_{ddX}} = {m_{{M_2}{O_n}}} + {m_{dd{H_2}S{O_4}}}$ ⇒${m_{ddX}} = \dfrac{a}{2}(2M + 16n) + 133,33an = a(M + 141,33n)$

$\begin{gathered}
  C{\% _{{M_2}{{(S{O_4})}_n}}} = \dfrac{{\dfrac{a}{2}(2M + 96n)}}{{a(M + 141,33n)}}.100\%  = 41,67\%  \hfill \\
   \Leftrightarrow M + 48n = 0,4167(M + 141,33n) \hfill \\
   \Rightarrow M = 18,67n \hfill \\
   \Rightarrow \left\{ \begin{gathered}
  n = 3 \hfill \\
  M = 56 \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} $

M là $Fe$

$\begin{gathered}
  {n_{FeS}} = \dfrac{{4,4}}{{88}} = 0,05mol \Rightarrow {n_{F{e_2}{O_3}}} = {n_{F{e_2}{{(S{O_4})}_3}}} = 0,025mol \hfill \\
  {n_{{H_2}S{O_4}}} = 3{n_{F{e_2}{O_3}}} = 0,075mol \hfill \\
   \Rightarrow {m_{F{e_2}{O_3}}} = 0,025.160 = 4g;{m_{F{e_2}{{(S{O_4})}_3}}} = 0,025.400 = 10g \hfill \\
  {m_{dd{H_2}S{O_4}}} = \dfrac{{0,075.98}}{{36,75\% }} = 20g \hfill \\
  {m_{ddX}} = {m_{F{e_2}{O_3}}} + {m_{dd{H_2}S{O_4}}} = 24g \hfill \\ 
\end{gathered} $

Gọi CTPT của Y là $F{e_2}{(S{O_4})_3}.n{H_2}O$

$\begin{gathered}
  {m_{ddsau}} = {m_{ddX}} - {m_Y} = 24 - 5,62 = 18,38g \hfill \\
  {m_{F{e_2}{{(S{O_4})}_3}sau}} = \dfrac{{{m_{ddsau}}.32,64}}{{100}} = 6g \hfill \\
   \Rightarrow {m_{F{e_2}{{(S{O_4})}_3}(Y)}} = 10 - 6 = 4g \hfill \\
   \Rightarrow {n_{F{e_2}{{(S{O_4})}_3}}} = 0,01mol \hfill \\
   \Rightarrow {n_Y} = {n_{F{e_2}{{(S{O_4})}_3}}} = 0,01mol \hfill \\
   \Rightarrow {M_Y} = \dfrac{{5,62}}{{0,01}} = 562 \Leftrightarrow 400 + 18n = 562 \hfill \\
   \Rightarrow n = 9 \hfill \\ 
\end{gathered} $

CTPT của Y: $F{e_2}{(S{O_4})_3}.9{H_2}O$