Hướng dẫn:

${{n}_{NaOH}}={{V}_{NaOH}}.{{C}_{{{M}_{NaOH}}}}=0,1.2=0,2\left( mol \right)$

${{m}_{NaOH}}={{n}_{NaOH}}.{{M}_{NaOH}}=0,2.40=8\left( g \right)$

${{\left( RCOO \right)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\xrightarrow{\text{ax}it,{{t}^{0}}}{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}+3RCOOH$

Bảo toàn khối lượng có:

${{m}_{CB}}+{{m}_{NaOH}}={{m}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}+{{m}_{RCOOH}}$

${{m}_{CB}}={{m}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}+{{m}_{RCOOH}}-{{m}_{NaOH}}$

$\Rightarrow {{m}_{CB}}=5,6+4,5-8=2,1\left( g \right)$

$\Rightarrow $Chọn đáp án B