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Câu hỏi: Đốt cháy hoàn toàn $m$ gam một chất béo triglixerit cần 1,61 mol ${{O}_{2}}$, sinh ra 1,14 mol $C{{O}_{2}}$ và 1,06 mol ${{H}_{2}}O$. Cho 7,088 gam chất béo tác dụng vừa đủ với dung dịch $NaOH$ thì khối lượng muối tạo thành là:
A. 7,612 gam.
B. 7,512 gam.
C. 7,412 gam.
D. 7,312 gsm.
A. 7,612 gam.
B. 7,512 gam.
C. 7,412 gam.
D. 7,312 gsm.
$\begin{aligned}
& \xrightarrow{BT.O}6{{n}_{X}}+2{{n}_{{{O}_{2}}}}=2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}\to {{n}_{X}}=\dfrac{1,14.2+1,06-2.1,61}{6}=0,02\ mol \\
& \xrightarrow{BTKL}{{m}_{X}}={{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}-{{m}_{{{O}_{2}}}}=17,72\ gam \\
\end{aligned}$
$\to $ Trong 7,088 gam $X$ có ${{n}_{X}}=\dfrac{7,088}{17,72}.0,02=0,008\ mol\to \left\{ \begin{aligned}
& {{n}_{NaOH}}=3{{n}_{X}}=0,024\ mol \\
& {{n}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=0,008\ mol \\
\end{aligned} \right.$
$\xrightarrow{BTKL}{{m}_{muoi}}={{m}_{X}}+{{m}_{NaOH}}-{{m}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=7,088+40.0,024-92.0,008=7,312\ gam$.
& \xrightarrow{BT.O}6{{n}_{X}}+2{{n}_{{{O}_{2}}}}=2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}\to {{n}_{X}}=\dfrac{1,14.2+1,06-2.1,61}{6}=0,02\ mol \\
& \xrightarrow{BTKL}{{m}_{X}}={{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}-{{m}_{{{O}_{2}}}}=17,72\ gam \\
\end{aligned}$
$\to $ Trong 7,088 gam $X$ có ${{n}_{X}}=\dfrac{7,088}{17,72}.0,02=0,008\ mol\to \left\{ \begin{aligned}
& {{n}_{NaOH}}=3{{n}_{X}}=0,024\ mol \\
& {{n}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=0,008\ mol \\
\end{aligned} \right.$
$\xrightarrow{BTKL}{{m}_{muoi}}={{m}_{X}}+{{m}_{NaOH}}-{{m}_{{{C}_{3}}{{H}_{5}}{{\left( OH \right)}_{3}}}}=7,088+40.0,024-92.0,008=7,312\ gam$.
Đáp án D.