\(\begin{array}{l} {{\rm{n}}_{COO}} = {n_{COONa}} = 2{n_{N{a_2}C{O_3}}} = 2.0,35 = 0,7{\rm{ mol}}\\ {\rm{ T}} \to \left| \begin{array}{l} {n_{C{O_2}}} = 0,35{\rm{ mol}} \to {n_{C{\rm{ (T)}}}} = {n_{N{a_2}C{O_3}}} + {n_{C{O_2}}} = 0,7 = {n_{COONa}}\\ \to T \Leftrightarrow \left\{ \begin{array}{l} HCOONa:0,3{\rm{ (BT}}{\rm{.H)}}\\ {(COONa)_2}:0,2 \end{array} \right. \end{array} \right.\\ {\rm{PP 3T }} \to \left\{ \begin{array}{l} COO:0,7\\ C{H_2}:x\\ {H_2}:y \end{array} \right. \to \left| \begin{array}{l} T\left\{ \begin{array}{l} HCOONa:0,3\\ {(COONa)_2}:0,2 \end{array} \right.\\ 28,6{\rm{ gam ancol}} \end{array} \right.\\ \to \left\{ \begin{array}{l} {n_{C{O_2}}} - {n_{{H_2}O}} \Leftrightarrow (x + 0,7) - (x + y) = 0,425\\ (44.0,7 + 14.x + 2y + 40.0,7 = (68.0,3 + 134.0,2) + 28,6 \end{array} \right. \to \left\{ \begin{array}{l} x = 1,175\\ y = 0,275 \end{array} \right.\\ - {\rm{ TH 1:}}{\rm{ E}}\left\{ \begin{array}{l} HCOOR':a\\ {(COOR')_2}:b\\ {(HCOO)_3}{C_3}{H_5}:c \end{array} \right. \to \left\{ \begin{array}{l} {n_E} = a + b + c = 0,275\\ {n_{HCOO}} = a + 3c = 0,3\\ {n_{{{(COONa)}_2}}} = b = 0,2 \end{array} \right. \to \left\{ \begin{array}{l} a = - 0,0375\\ \to Loai \end{array} \right.\\ - {\rm{ TH 2: E }}\left\{ \begin{array}{l} HCOOR':a\\ {(COOR')_2}:b\\ HCOOROOC - COO - R':c \end{array} \right. \to \left\{ \begin{array}{l} a + b + c = 0,275\\ a + c = 0,3\\ b + c = 0,2 \end{array} \right. \to \left\{ \begin{array}{l} b = - 0,025\\ \to Loi \end{array} \right.\\ - {\rm{ TH 3: E }}\left\{ \begin{array}{l} HCOOR':a\\ {(HCOO)_2}R'':b\\ HCOOROOC - COO - R':0,2 \end{array} \right. \to \left\{ \begin{array}{l} a + 2b + c = 0,3\\ a + b + c = 0,275 \end{array} \right. \to \left\{ \begin{array}{l} a = 0,05\\ b = 0,025 \end{array} \right.\\ \to 28,6{\rm{ gam}}\left\{ \begin{array}{l} R'OH:0,25\\ R''{(OH)_2}:0,225 \end{array} \right. \to 0,25.M + 0,225.M' = 28,6 \to \left\{ \begin{array}{l} M = 46{\rm{ (}}{{\rm{C}}_2}{H_5}OH)\\ M' = 76{\rm{ (}}{{\rm{C}}_3}{H_6}{(OH)_2} \end{array} \right.\\ \to {m_{HCOO{C_2}{H_5}}} = 74.0,05 = 3,7{\rm{ gam}} \end{array}\)