Đáp án:
\(\begin{array}{l}
{m_{Al}} = 2,7g\\
{m_{FeC{O_3}}} = 11,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
FeC{O_3} + {H_2}S{O_4} \to FeS{O_4} + C{O_2} + {H_2}O\\
{m_{hh}} - {m_{C{O_2}}} - {m_{{H_2}}} = 9,6g \Rightarrow {m_{C{O_2}}} + {m_{{H_2}}} = 4,7g\\
hh:Al(a\,mol),FeC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
27a + 116b = 14,3\\
1,5a \times 2 + 44b = 4,7
\end{array} \right.\\
\Rightarrow a = b = 0,1\,mol\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{FeC{O_3}}} = 0,1 \times 116 = 11,6g
\end{array}\)
