\(n_{Fe}=\frac{2,8}{56}=0,05\left(mol\right)\)
\(m_{HCl}=\frac{60.7,3\%}{100\%}=4,38\left(g\right)\)
\(n_{HCl}=\frac{4,38}{36,5}=0,12\left(mol\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ban đầu: 0,05___0,12
Phản ứng: 0,05____0,1_______0,05 (mol)
Dư:____________0,02
Lập tỉ lệ: \(\frac{0,05}{1}< \frac{0,12}{2}\left(0,05< 0,06\right)\)
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(\Rightarrow Fe\) hết \(HCl\) dư
\(m_{ddHCl\left(pư\right)}=\frac{3,65.100\%}{7,3\%}=50\left(g\right)\)
\(m_{ddX}=m_{Fe}+m_{ddHCl\left(pư\right)}-m_{H_2}=2,8+50-0,05.2=52,7\left(g\right)\)
\(C\%_{FeCl_2}=\frac{0,05.127}{52,7}.100\%=12,05\%\)
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