Đáp án:
b) %C: 10,79%; %S: 28,78%; %Fe: 60,43%
c) $F{e_3}{O_4}$
Giải thích các bước giải:
a)
$\begin{gathered}
2xFe + y{O_2}\xrightarrow{{{t^o}}}2F{e_x}{O_y} \hfill \\
S + {O_2}\xrightarrow{{{t^o}}}S{O_2} \hfill \\
C + {O_2}\xrightarrow{{{t^o}}}C{O_2} \hfill \\
S{O_2} + Ca{(OH)_2} \to CaS{O_3} + {H_2}O \hfill \\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O \hfill \\
\end{gathered} $
b)
Gọi số mol $C$, $S$ là a, b
$\begin{gathered}
{n_{hhkhi}} = \dfrac{{13,44}}{{22,4}} = 0,6mol;{n_{{O_2}du}} = \dfrac{{2,24}}{{22,4}} = 0,1mol \hfill \\
\Rightarrow {n_{C{O_2}}} + {n_{S{O_2}}} = 0,6 - 0,1 = 0,5 \Leftrightarrow a + b = 0,5{\text{ (1)}} \hfill \\
\end{gathered} $
$\begin{gathered}
{m_ \downarrow } = {m_{CaS{O_3}}} + {m_{CaC{O_3}}} = 55\left( g \right) \hfill \\
\Leftrightarrow 100a + 120b = 55{\text{ (2)}} \hfill \\
\end{gathered} $
Từ (1) và (2) ⇒ x = y = 0,25
$\begin{gathered}
\% {m_C} = \dfrac{{0,25.12}}{{27,8}}.100\% = 10,79\% \hfill \\
\% {m_S} = \dfrac{{0,25.32}}{{27,8}}.100\% = 28,78\% \hfill \\
\% {m_{Fe}} = 100 - 10,79 - 28,78 = 60,43\% \hfill \\
\end{gathered} $
c)
$\begin{gathered}
{m_{Fe}} = 27,8 - {m_C} - {m_S} = 27,8 - 3 - 8 = 16,8 \hfill \\
\Rightarrow {n_{Fe}} = 0,3mol \hfill \\
{m_{O(X)}} = {m_X} - {m_{Fe}} = 23,2 - 16,8 = 6,4 \hfill \\
\Rightarrow {n_{O(X)}} = 0,4 \hfill \\
\Rightarrow \dfrac{x}{y} = \dfrac{{{n_{Fe}}}}{{{n_O}}} = \dfrac{3}{4} \hfill \\
\Rightarrow X:F{e_3}{O_4} \hfill \\
\end{gathered} $
