a)

$\begin{gathered} 3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4} \hfill \\ C + {O_2}\xrightarrow{{{t^o}}}C{O_2} \hfill \\ S + {O_2}\xrightarrow{{{t^o}}}S{O_2} \hfill \\ \end{gathered} $

b)

$\begin{gathered} {n_{F{e_3}{O_4}}} = \dfrac{{23,2}}{{232}} = 0,1{\text{ mol}} \Rightarrow {n_{Fe}} = 0,3{\text{ mol}} \Rightarrow {m_{Fe}} = 16,8\left( g \right) \hfill \\ \Rightarrow \% {m_{Fe}} = \dfrac{{16,8}}{{27,8}}.100\% = 60,43\% \hfill \\ \end{gathered} $

${V_{{O_2}du}} = 2,24\left( l \right) \Rightarrow {V_{C{O_2}}} + {V_{S{O_2}}} = 13,44 - 2,24 = 11,2\left( l \right)$

Gọi x và y lần lượt là số mol C và S

Ta có hpt: $\left\{ \begin{gathered} 12x + 32y = 27,8 - 16,8 \hfill \\ x + y = \dfrac{{11,2}}{{22,4}} = 0,5 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x = 0,25 \hfill \\ y = 0,25 \hfill \\ \end{gathered} \right.$

$\begin{gathered} \Rightarrow \% {m_C} = \dfrac{{0,25.12}}{{27,8}}.100\% = 10,79\% \hfill \\ \Rightarrow \% {m_S} = 100 - 10,79 - 60,43 = 28,78\% \hfill \\ \end{gathered} $

c) Bảo toàn nguyên tố O:

${n_{{O_2}(bd)}} = {n_{C{O_2}}} + {n_{S{O_2}}} + {n_{{O_2}(du)}} = 0,25 + 0,25 + 0,1 = 0,6 \\\Rightarrow V = 13,44\left( l \right)$

${m_{kt}} = {m_{CaC{O_3}}} + {m_{CaS{O_3}}} = 0,25.100 + 0,25.120 = 55\left( g \right)$