\(n_{NaOH}=\dfrac{200.12\%}{40}=0,6\left(mol\right)\\ 3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\\ 2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ A:Fe\left(OH\right)_3;B:NaCl\\ a,n_{Fe\left(OH\right)_3}=\dfrac{0,6}{3}=0,2\left(mol\right);n_{NaCl}=n_{NaOH}=0,6\left(mol\right)\\ m_A=m_{Fe\left(OH\right)_3}=107.0,2=21,4\left(g\right)\\ m_{ddB}=200+200-21,4=378,6\left(g\right)\\ C\%_{ddNaCl}=\dfrac{0,6.58,5}{378,6}.100\approx9,271\%\\ b,Đặt:x=n_{Fe\left(OH\right)_3\left(p.ứ\right)}=x\left(mol\right)\\ \Rightarrow m_{rắn}=m_{Fe\left(OH\right)_3\left(chưa.phân.huỷ\right)}+m_{Fe_2O_3}=\left(21,4-107x\right)+80x=16,08\\ \Leftrightarrow21,4-27x=16,08\\ \Leftrightarrow27x=5,32\Leftrightarrow x=\dfrac{5,32}{27}\Rightarrow\%m_{A\left(nung\right)}=\dfrac{\dfrac{5,32}{27}}{0,2}.100\approx98,519\%\)
