Đáp án đúng: B
Chọn a = 240 gam.
$\Rightarrow {{n}_{{{H}_{2}}}}=\frac{11}{2}=5,5\,mol;\,\,{{n}_{C{{H}_{3}}COOH}}=\frac{240.C\%}{60}=0,04C\,mol;\,\,{{n}_{{{H}_{2}}O}}=\,\frac{240-2,4C}{18}mol$
Phương trình phản ứng
$2C{{H}_{3}}COOH\,+\,2NaOH\,\to \,2C{{H}_{3}}{COONa}\,{+}\,{{{H}}_{2}}\,\,\,\,(1)$
$2{{H}_{2}}O\,+\,2Na\,\to \,2NaOH\,+\,{{H}_{2}}\,\,\,\,(2)$
Từ (1), (2) suy ra
${{n}_{C{{H}_{3}}COOH}}+{{n}_{{{H}_{2}}O}}=2{{n}_{{{H}_{2}}}}\Rightarrow 0,04C+\frac{240-2,4C}{18}=2.5,5\Rightarrow C=25$