Đáp án đúng: D
Phương pháp giải:
\(X\left\{ \begin{array}{l}Al:0,5\\Fe\\O\end{array} \right. \to Y\left\{ \begin{array}{l}Al\\A{l_2}{O_3}\\Fe\\O\end{array} \right. + \left[ \begin{array}{l}NaOH \to \left\{ \begin{array}{l}{H_2}:0,15\\46,72(g)\left\{ \begin{array}{l}Fe\\O\end{array} \right.\end{array} \right.\\HCl \to 158,19(g)\left\{ \begin{array}{l}AlC{l_3}:0,5\\FeC{l_2}\end{array} \right.\end{array} \right.\)BTNT Al: nAl (X) = nAlCl3 Ta có: mT = mAlCl3 + mFeCl2 ⟹ nFeCl2 = nFeTrong 46,72 gam chất rắn có: 46,72 = mFe + mO ⟹ nO Ta có: Al + NaOH + H2O ⟶ NaAlO2 + 3/2 H2Từ nH2 ⟹ nAlBTNT Al: nAl (X) = nAl + 2.nAl2O3 ⟹ nAl2O3 = 0,2 molBTNT O: nO (X) = 3.nAl2O3 + nO Ta có: nFe2O3 = x; nFe3O4 = yBTNT Fe, BTNT O ⟹ x , y ⟹ x : y Giải chi tiết:\(X\left\{ \begin{array}{l}Al:0,5\\Fe\\O\end{array} \right. \to Y\left\{ \begin{array}{l}Al\\A{l_2}{O_3}\\Fe\\O\end{array} \right. + \left[ \begin{array}{l}NaOH \to \left\{ \begin{array}{l}{H_2}:0,15\\46,72(g)\left\{ \begin{array}{l}Fe\\O\end{array} \right.\end{array} \right.\\HCl \to 158,19(g)\left\{ \begin{array}{l}AlC{l_3}:0,5\\FeC{l_2}\end{array} \right.\end{array} \right.\)BTNT Al: nAl (X) = nAlCl3 = 0,5 molTa có: mT = mAlCl3 + mFeCl2 ⇔ 158,19 = 0,5.133,5 + 127.nFeCl2 ⟹ nFeCl2 = 0,72 mol = nFeTrong 46,72 gam chất rắn có: 46,72 = mFe + mO ⇔ 46,72 = 0,72.56 + 16.nO ⟹ nO = 0,4 molTa có: nH2 = 3,36/22,4 = 0,15 molAl + NaOH + H2O ⟶ NaAlO2 + 3/2 H20,1 ⟵ 0,15BTNT Al: nAl (X) = nAl + 2.nAl2O3 ⇔ 0,5 = 0,1 + 2.nAl2O3 ⟹ nAl2O3 = 0,2 molBTNT O: nO (X) = 3.nAl2O3 + nO = 3.0,2 + 0,4 = 1 molTa có: nFe2O3 = x; nFe3O4 = yBTNT Fe: 2x + 3y = 0,72BTNT O: 3x + 4y = 1⟹ x = 0,12; y = 0,16 ⟹ x : y = 3:4
