Ta có: \(\left\{{}\begin{matrix}n_{FeCl_3}=0,1\left(mol\right)\\n_{CuCl_2}=0,15\left(mol\right)\end{matrix}\right.\)
\(n_{Fe}=0,25\left(mol\right)\)
\(Fe\left(0,05\right)+2FeCl_3\left(0,1\right)\rightarrow3FeCl_2\)
\(Fe\left(0,15\right)+CuCl_2\left(0,15\right)\rightarrow FeCl_2+Cu\left(0,15\right)\)
\(\Rightarrow C.rănA:\left\{{}\begin{matrix}Fe\left(dư\right)=0,25-0,05-0,15=0,05\left(mol\right)\\n_{Cu}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_A=0,05.56+0,15.64=12,4\left(g\right)\)
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