Câu hỏi
Xà phòng hóa hoàn toàn m gam chất béo trung tính bằng dung dịch KOH thu được 18,77 gam muối. Nếu thay dung dịch KOH bằng dung dịch NaOH thì chỉ thu được 17,81 gam muối. Giá trị của m là
- A 18,36.
- B 17,25.
- C 17,65.
- D 36,58.
Phương pháp giải:
Giả sử chất béo là \({(\overline R C{\text{OO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}\)
\(\begin{gathered}
{(\overline R C{\text{OO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}\xrightarrow{{ + KOH}}3\overline R C{\text{OO}}K \hfill \\
{(\overline R C{\text{OO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}\xrightarrow{{ + NaOH}}3\overline R C{\text{OONa}} \hfill \\
\,\,\,\,\,\,\,\,\overline R C{\text{OO}}K..............\overline R C{\text{OONa}} \hfill \\
PT:\overline R + 83..................\overline R + 67 \hfill \\
DB:18,77g................17,81g \hfill \\
\to 18,77(\overline R + 67) = 17,81(\overline R + 83) \hfill \\
\to \overline R \hfill \\
\end{gathered} \)
Lời giải chi tiết:
Giả sử chất béo là \({(\overline R C{\text{OO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}\)
\(\begin{gathered}
{(\overline R C{\text{OO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}\xrightarrow{{ + KOH}}3\overline R C{\text{OO}}K \hfill \\
{(\overline R C{\text{OO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}}\xrightarrow{{ + NaOH}}3\overline R C{\text{OONa}} \hfill \\
\,\,\,\,\,\,\,\,\overline R C{\text{OO}}K..............\overline R C{\text{OONa}} \hfill \\
PT:\overline R + 83..................\overline R + 67 \hfill \\
DB:18,77g................17,81g \hfill \\
\to 18,77(\overline R + 67) = 17,81(\overline R + 83) \hfill \\
\to \overline R = \frac{{1379}}{6} \hfill \\
n\overline R C{\text{OO}}K = \frac{{18,77}}{{\frac{{1379}}{6} + 83}} = 0,06(mol) \hfill \\
\to n{(\overline R C{\text{OO)}}_{\text{3}}}{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{5}}} = 0,02mol \hfill \\
\to m = 0,02\left[ {(\frac{{1379}}{6} + 44).3 + 41} \right] = 17,25(g) \hfill \\
\end{gathered} \)
Đáp án B