Lời giải:Đáp án B
\({{n}_{C{{O}_{2}}}}:2,84\,(mol);\,\,\,{{n}_{{{H}_{2}}O}}:2,74\,\,(mol);\,\,{{n}_{{{O}_{2}}(\tilde{n}o\acute{a}t)}}:x\,(mol);\,\,{{n}_{{{N}_{2}}}}:y\,(mol)\)
Sơ đồ phản ứng đốt cháy: \(X+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O+{{N}_{2}}\)
\({{n}_{{{O}_{2}}}}=1,5.({{n}_{C{{O}_{2}}}}-{{n}_{{{N}_{2}}}})\leftrightarrow x=1,5(2,84-y)\,\,(1)\)
\(\xrightarrow{BTO}{{n}_{O\,\,(trong\,\,X)}}=2.2,84+2,74-2x=8,42-2x\)
Phản ứng thuỷ phân: \(X+NaOH\to RN{{O}_{2}}Na+{{H}_{2}}O\)
\(\xrightarrow{BTN}{{n}_{NaOH}}=2y\)
\(\xrightarrow{BTO}{{n}_{{{H}_{2}}O}}={{n}_{O\,(X)}}+{{n}_{O\,\,(NaOH)}}-{{n}_{(O)RN{{O}_{2}}Na}}=(8,42-2x)+2y-4y=8,42-2x-2y\)
\(\xrightarrow{BTKL}m+2y.40=m+29,7+18.(8,42-2x-2y)\,\,(2)\)
Từ (1)(2) \(\to x=3,585;\,\,y=0,45\to m=72,16\,\,(gam)\)
