\(n_{Fe}=\dfrac{4.48}{56}=0.08\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.08....0.16.........0.08.....0.08\)
\(m_{HCl}=0.16\cdot36.5=5.84\left(g\right)\)
\(C\%_{HCl}=\dfrac{5.84}{800}\cdot100\%=0.73\%\)
\(m_{FeCl_2}=0.08\cdot127=10.16\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=4.48+800-0.08\cdot2=804.32\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{10.16}{804.32}\cdot100\%=1.26\%\)
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