\(n_{FeCl_2}=\dfrac{50,8}{127}=0,4\) mol
Đặt \(n_{FeCl_3}=x\) mol
\(\left(Cl\right):n_{HCl}=0,4.2+3.x=0,8+3x\) mol
\(\left(H\right):n_{H_2O}.2=n_{HCl}=>n_{H_2O}=0,4+1,5x\) mol
\(m_x+m_{HCl}=m_{muối}+m_{H_2O}\)
\(=>10,8+\left(0,8+3x\right).36,5=50,8+162,5x+\left(0,4+1,5x\right).18\)
\(=>x=1=>m_{FeCl_3}=1.162,5=162,5\) g