Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}T{N_1}:0,05\,mol\,E\left\{ \begin{array}{l}CONH:x\\C{H_2}:y\\{H_2}O:0,05\end{array} \right. + {O_2} \to \left\{ \begin{array}{l}C{O_2}:x + y(BT:C)\\{H_2}O:0,5x + y + 0,05(BT:H)\end{array} \right.\\{n_{C{O_2}}} - {n_{{H_2}O}} = 0,045 \to (x + y) - (0,5x + y + 0,05) = 0,045 \to x = 0,19(mol)\\ \to {m_E} = 43.0,19 + 14y + 0,05.18 = 14y + 9,07(g)\\Ty\,le:\,\,\,\,\,\,\,E\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,NaOH\\\,\,\,\,\,\,\,\,\,\,14y + 9,07(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,19(mol)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,119,6\,\,\,\,\,\,(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,52(mol)\\ \to 1,52(14y + 9,07) = 0,19.119,6 \to y = 0,42(mol)\\14,95(g)E\left\{ \begin{array}{l}CONH:0,19\\C{H_2}:0,42\\{H_2}O:0,05\end{array} \right. \to 21,65(g)\,\,muoi\left\{ \begin{array}{l}COONa:0,19\\N{H_2}:0,19\\C{H_2}:0,42\end{array} \right.\\119,6(g)E\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 173,2(g)\end{array}\)
Đáp án D