Question

Answer 1

a) Gọi số mol FeO, Al₂O₃ là a, b (mol)

=> 72a + 102b = 21 (1)

\(n_{HCl}=\dfrac{200.16,425\%}{36,5}=0,9\left(mol\right)\)

PTHH: FeO + 2HCl --> FeCl₂ + H₂O

               a---->2a-------->a

             Al₂O₃ + 6HCl --> 2AlCl₃ + 3H₂O

                b------>6b------>2b

=> 2a + 6b = 0,9 (2)

(1)(2) => a = 0,15 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}\%m_{FeO}=\dfrac{0,15.72}{21}.100\%=51,43\%\\\%m_{Al_2O_3}=\dfrac{0,1.102}{21}.100\%=48,57\%\end{matrix}\right.\)

b) m_{dd sau pư} = 21 + 200 = 221 (g)

m_FeCl2 = 0,15.127 = 19,05 (g)

m_AlCl3 = 0,2.133,5 = 26,7 (g)

\(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{19,05}{221}.100\%=8,62\%\\C\%_{AlCl_3}=\dfrac{26,7}{221}.100\%=12,08\%\end{matrix}\right.\)