Hỗn hợp E gồm chất X (CmH2m+4

C

- Hướng dẫn giải

Phương pháp giải:

\(\begin{gathered}
0,2mol\,E\left\{ \begin{gathered}
X:{C_m}{H_2}_{m + 4}{O_4}{N_2}(xmol) \hfill \\
Y:{C_n}{H_2}_{n + 3}{O_2}N(ymol) \hfill \\
\end{gathered} \right. \hfill \\
\to x + y = 0,2(1) \hfill \\
\xrightarrow{{BTH}}x(2m + 4) + y(2n + 3) = 0,84.2 \to 2mx + 2ny + 4x + 3y = 1,68(2) \hfill \\
\xrightarrow{{BTC}}{n_{C{O_2}}} = mx + ny \hfill \\
\xrightarrow{{BTO}}4x + 2y + 0,58.2 = 2mx + 2ny + 0,84 \to 2mx + 2ny - 4x - 2y = 0,32(3) \hfill \\
\xrightarrow{{(1)(2)(3)}}\left\{ \begin{gathered}
2mx + 2ny = ? \hfill \\
x = ? \hfill \\
y = ? \hfill \\
\end{gathered} \right. \to 0,24m + 0,16n = 0,96 \to 3m + 2n = 12 \hfill \\
\to \left\{ \begin{gathered}
m = ? \hfill \\
n = ? \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \)

Giải chi tiết:

\(\begin{gathered}
0,2mol\,E\left\{ \begin{gathered}
X:{C_m}{H_2}_{m + 4}{O_4}{N_2}(xmol) \hfill \\
Y:{C_n}{H_2}_{n + 3}{O_2}N(ymol) \hfill \\
\end{gathered} \right. \hfill \\
\to x + y = 0,2(1) \hfill \\
\xrightarrow{{BTH}}x(2m + 4) + y(2n + 3) = 0,84.2 \to 2mx + 2ny + 4x + 3y = 1,68(2) \hfill \\
\xrightarrow{{BTC}}{n_{C{O_2}}} = mx + ny \hfill \\
\xrightarrow{{BTO}}4x + 2y + 0,58.2 = 2mx + 2ny + 0,84 \to 2mx + 2ny - 4x - 2y = 0,32(3) \hfill \\
\xrightarrow{{(1)(2)(3)}}\left\{ \begin{gathered}
2mx + 2ny = 0,96 \hfill \\
x = 0,12 \hfill \\
y = 0,08 \hfill \\
\end{gathered} \right. \to 0,24m + 0,16n = 0,96 \to 3m + 2n = 12 \hfill \\
\to \left\{ \begin{gathered}
m = 2 \hfill \\
n = 3 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
X:{(COON{H_4})_2} \hfill \\
Y:{C_2}{H_5}COON{H_4} \hfill \\
\end{gathered} \right. \to Muoi\left\{ \begin{gathered}
{(C{\text{OONa)}}_{\text{2}}}{\text{:0,12}} \hfill \\
{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{COONa:0,08}} \hfill \\
\end{gathered} \right. \hfill \\
\to a = 0,12.134 + 0,08.96 = 23,76(g) \hfill \\
\end{gathered} \)

Đáp án C