Đáp án:
$FeS{O_4}.7{H_2}O$
Giải thích các bước giải:
+ ở ${80^o}C$:
$\begin{gathered}
S = \dfrac{{{m_{ct}}}}{{{m_{{H_2}O}}}}.100 = \dfrac{{{m_{ct}}}}{{75,875 - {m_{ct}}}}.100 = 68,3 \hfill \\
\Rightarrow {m_{XS{O_4}(1)}} = 30,792\left( g \right) \hfill \\
\end{gathered} $
+ Ở ${20^o}C$:
$\begin{gathered}
{m_{ddsau}} = 75,875 - 27,8 = 48,075\left( g \right) \hfill \\
S = \dfrac{{{m_{ct}}}}{{{m_{{H_2}O}}}}.100 = \dfrac{{{m_{ct}}}}{{48,075 - {m_{ct}}}}.100 = 48 \hfill \\
\Rightarrow {m_{XS{O_4}(2)}} = 15,592\left( g \right) \hfill \\
\end{gathered} $
$\begin{gathered}
\Rightarrow {m_{XS{O_4}(tt)}} = {m_{XS{O_4}(1)}} - {m_{XS{O_4}(2)}} = 30,792 - 15,592 = 15,2\left( g \right) \hfill \\
\Rightarrow {m_{{H_2}O(tt)}} = 27,8 - 15,2 = 12,6\left( g \right) \hfill \\
\Rightarrow {n_{tt}} = {n_{XS{O_4}(tt)}} = \dfrac{1}{7}{n_{{H_2}O}} = \dfrac{1}{7}.\dfrac{{12,6}}{{18}} = 0,1mol \hfill \\
\Rightarrow {M_{XS{O_4}}} = \dfrac{{15,2}}{{0,1}} = 152 \Leftrightarrow X + 96 = 152 \Rightarrow X = 56 \hfill \\
\end{gathered} $
⇒ X là $Fe$
Tinh thể là $FeS{O_4}.7{H_2}O$
