Đáp án đúng: D
Phương pháp giải:
\(34,24(g)X\left\{ \begin{array}{l}FeC{O_3}:a\\F{e_3}{O_4}:b\\Fe{(N{O_3})_2}:c\end{array} \right. + \left\{ \begin{array}{l}NaN{O_3}\\NaHS{O_4}:d\end{array} \right. \to \left\{ \begin{array}{l}Z:\left\{ \begin{array}{l}C{O_2}:a\\NO:3a\end{array} \right.\\Y\left\{ \begin{array}{l}F{e^{3 + }}\\N{a^ + }\\NO_3^ - \\SO_4^{2 - }\end{array} \right. + \left[ \begin{array}{l}Fe:0,15\\Ba{(OH)_2} \to 209,18(g) \downarrow \end{array} \right.\end{array} \right.\)Gọi nFeCO3 = a mol; nFe3O4 = b mol; nFe(NO3)2 = c mol⟹ nCO2 ; nNO mX = mFeCO3 + mFe3O4 + mFe(NO3)2 (1)BT e ⟹ 2nFe = 2nFe3O4 + 3nNO (2) Đặt: nNaHSO4 = d mol⟹ nH+ = 2nCO2 + 2nO (Oxit) + 4nNO (3)BTNT Fe ⟹ nFe2+ ⟹ mkết tủa = mBaSO4 + mFe(OH)3 + mFe(OH)2 (4)Từ (1), (2), (3), (4) ⟹ a; b; c; d ⟹ %mFe3O4 Giải chi tiết:\(34,24(g)X\left\{ \begin{array}{l}FeC{O_3}:a\\F{e_3}{O_4}:b\\Fe{(N{O_3})_2}:c\end{array} \right. + \left\{ \begin{array}{l}NaN{O_3}\\NaHS{O_4}:d\end{array} \right. \to \left\{ \begin{array}{l}Z:\left\{ \begin{array}{l}C{O_2}:a\\NO:3a\end{array} \right.\\Y\left\{ \begin{array}{l}F{e^{3 + }}\\N{a^ + }\\NO_3^ - \\SO_4^{2 - }\end{array} \right. + \left[ \begin{array}{l}Fe:0,15\\Ba{(OH)_2} \to 209,18(g) \downarrow \end{array} \right.\end{array} \right.\)Gọi nFeCO3 = a mol; nFe3O4 = b mol; nFe(NO3)2 = c mol⟹ 116a + 232b + 180c = 34,24 (1)⟹ nCO2 = a mol; nNO = 3a mol.Cho Fe vào dung dịch Y không thấy khí thoát ra ⟹ H+ hết.nFe = 8,4/56 = 0,15 mol.BT electron: 0,15.2 = 2b + 3.3c (2) Đặt: nNaHSO4 = d mol⟹ nH+ = 2nCO2 + 2nO (Oxit) + 4nNO ⇔ d = 2a + 2.4.b + 4.3a = 14a + 8b (3)nFe3+ = 2nFe = 2.0,15 = 0,3BTNT Fe: nFe2+ = a + 3b + c - 0,3⟹ mkết tủa = mBaSO4 + mFe(OH)3 + mFe(OH)2 ⇔ 209,18 = 233.d + 107.0,3 + 90.(a + 3b + c - 0,3) (4)Từ (1), (2), (3), (4) ⟹ a = 0,02; b = 0,06; c = 0,1; d = 0,76⟹ %mFe3O4 = 232.0,06/34,24 = 40,65%
