Phương pháp giải:

Quy đổi hỗn hợp về Fe, C, O

Bảo toàn khối lượng, bảo toàn electron, bảo toàn nguyên tố

Lời giải chi tiết:

\(\left\{ \begin{gathered}
Fe:6x \hfill \\
F{e_3}{O_4}:x \hfill \\
FeC{O_3}:2x \hfill \\
\end{gathered} \right.\xrightarrow{{quy\,doi}}\left\{ \begin{gathered}
Fe:11x \hfill \\
C:10x \hfill \\
O:2x \hfill \\
\end{gathered} \right. + {H_2}S{O_4}\xrightarrow{{}}\left\langle \begin{gathered}
\left\{ \begin{gathered}
F{e^{2 + }},F{e^{3 + }} \hfill \\
S{O_4}^{2 - } \hfill \\
\end{gathered} \right.\xrightarrow{{0,2m\,Cu}}\left\{ \begin{gathered}
F{e^{2 + }},C{u^{2 + }} \hfill \\
S{O_4}^{2 - } \hfill \\
\end{gathered} \right. \hfill \\
0,095\,mol\,Z\left\{ \begin{gathered}
C{O_2}:2x(mol) \hfill \\
S{O_2}:\,z\,(mol) \hfill \\
\end{gathered} \right.\xrightarrow{{ + Ca{{(OH)}_2}}}a(g) \downarrow \left\{ \begin{gathered}
CaC{O_3}:\,2x(mol) \hfill \\
CaS{O_3}:\,z\,(mol) \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \right.\)

Quá trình nhường electron .        Quá trình nhận electron

\(\begin{gathered}
F{e^0}\,\,\,\, - 2e \to F{e^{ + 2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{S^{ + 6}} + 2e \to {S^{ + 4}} \hfill \\
11x \to 22x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2z \leftarrow z\,\,\,\,\,\,\,\,(mol) \hfill \\
{C^0} - 4e \to {C^{ + 4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{O^0}\,\,\,\,\,\,\, + 2e \to {O^{ - 2}} \hfill \\
2x \to 8x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10x \to 20x\,\,\,\,\,\,\,\,\,\,\,\,(mol) \hfill \\
C{u^0} - 2e \to C{u^{ + 2}} \hfill \\
y\,\,\,\,\,\, \to 2y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol)\, \hfill \\
\end{gathered} \)

\(\left\{ \begin{gathered}
{m_{hhE}} = 800x = m\,\,(1) \hfill \\
{m_{Cu}} = 64y = 0,2m\,(2) \hfill \\
{n_{hhZ}} = 2x + y = 0,095 \hfill \\
\xrightarrow{{BT:e}}22x + 8x + 2y = 2z + 20x \hfill \\
\end{gathered} \right. = > \left\{ \begin{gathered}
\xrightarrow{{(1):\,(2)}}5x = 2y \hfill \\
2x + y = 0,095 \hfill \\
10x + 2y - 2z = 0 \hfill \\
\end{gathered} \right. = > \left\{ \begin{gathered}
x = 0,01\,(mol) \hfill \\
y = 0,025\,(mol) \hfill \\
z = 0,075\,(mol) \hfill \\
\end{gathered} \right.\)

 => n_CaCO3 = 2x= 0,02 (mol)

n_CaSO3 = z = 0,075 (mol)

=> a = m_CaCO3 + m_CaSO3 = 0,02.100 + 0,075.120 = 11 (g)

Đáp án A