Đốt cháy hoàn toàn 44,4 gam hỗn hợp X gồm ba axit:...

B

- Hướng dẫn giải

Phương pháp giải:

Gọi số mol của CH₃COOH : x mol, C_nH_mCOOH:  y mol và HOOC-COOH : z mol

CH₃COOH + NaHCO₃ → CH₃COONa + CO₂ + H₂O          (1)

    x                                                        x

C_nH_mCOOH + NaHCO₃→ C_nH_mCOONa + CO₂ + H₂O      (2)

    y                                                                 y                                                                         

HOOC-COOH + 2NaHCO₃→NaOOC-COONa + 2CO₂ + 2H₂O     (3)

z                                                                       2z

=>\({n_{C{O_2}}} = x + y + 2z = \frac{{16,8}}{{22,4}} = 0,75\,mol\)

\({{\text{n}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}{\text{ = }}\frac{{{\text{21,6}}}}{{{\text{18}}}}{\text{ = 1,2}}\,{\text{mol}}\, \Rightarrow {{\text{n}}_{\text{H}}}{\text{ = }}\,{\text{2}}{{\text{n}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}{\text{ = }}\,{\text{2,4}}\,{\text{mol}}\)

- Bảo toàn nguyên tố oxi:

ĐLBTKL:

\(\begin{gathered}
{m_X} = {m_C} + {m_H} + {m_O} = 44,4\,gam \hfill \\
\Rightarrow \,{n_C} = ?\,mol \hfill \\
\Rightarrow \,{n_{C{O_2}}} = {n_C} = ?mol \hfill \\
\end{gathered} \)

Giải chi tiết:

Gọi số mol của CH₃COOH : x mol, C_nH_mCOOH:  y mol và HOOC-COOH : z mol

CH₃COOH + NaHCO₃CH₃COONa + CO₂ + H₂O          (1)

    x                                                        x

C_nH_mCOOH + NaHCO₃C_nH_mCOONa + CO₂ + H₂O      (2)

    y                                                                 y                                                                         

HOOC-COOH + 2NaHCO₃NaOOC-COONa + 2CO₂ + 2H₂O     (3)

z                                                                       2z

=>\({n_{C{O_2}}} = x + y + 2z = \frac{{16,8}}{{22,4}} = 0,75\,mol\)

\({{\text{n}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}{\text{ = }}\frac{{{\text{21,6}}}}{{{\text{18}}}}{\text{ = 1,2}}\,{\text{mol}}\, \Rightarrow {{\text{n}}_{\text{H}}}{\text{ = }}\,{\text{2}}{{\text{n}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}{\text{ = }}\,{\text{2,4}}\,{\text{mol}}\)

- Bảo toàn nguyên tố oxi:

\(\begin{gathered}
{{\text{n}}_{\text{O}}}{\text{ = 2}}{{\text{n}}_{{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}}}{\text{ + 2}}{{\text{n}}_{{{\text{C}}_{\text{n}}}{{\text{H}}_{\text{m}}}{\text{COOH}}}}{\text{ + 4}}{{\text{n}}_{{\text{HOOC - COOH}}}} \hfill \\
\,\,\,\,\,{\text{ = 2x + 2y + 4z}} \hfill \\
\,\,\,\,\,{\text{ = 2}}{\text{.0,75 = 1,5}}\,{\text{mol}} \hfill \\
\end{gathered} \)

- ĐLBTKL:

\(\begin{gathered}
{m_X} = {m_C} + {m_H} + {m_O} = 44,4\,gam \hfill \\
\Rightarrow \,{n_C} = \frac{{44,4 - 2,4 - 1,5.16}}{{12}} = 1,5\,mol \hfill \\
\Rightarrow \,{n_{C{O_2}}} = {n_C} = 1,5\,mol \hfill \\
\end{gathered} \)

=> a = 1,5.44= 66 gam

Đáp án B