Đốt cháy hoàn toàn 0,25 mol hỗn hợp X gồm hai hidr...

A

- Hướng dẫn giải

\(\left| \begin{array}{l}2{n_{C{O_2}}} + {n_{{H_2}O}} = 2{n_{{O_2}}} = 2.0,625\\30 - 44{n_{C{O_2}}} - 18{n_{{H_2}O}} = 4,3\end{array} \right. \Rightarrow \left| \begin{array}{l}{n_{C{O_2}}} = 0,4mol\\{n_{{H_2}O}} = 0,45mol\end{array} \right.\)

\(\left\{ \begin{array}{l}\overline C = \frac{{0,4}}{{0,25}} = 1,6 \to {n_{C{H_4}}}:{a^{mol}}\\\overline H = \frac{{0,45.2}}{{0,25}} = 3,6{ \to _{{C_n}{H_2}}}:{b^{mol}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a + b = 0,25\\4a + 2b = 0,45.2\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 0,2\\b = 0,05\end{array} \right.\)

Bảo toàn C: 0,2.1 + 0,05.n = 0,4 ⇒ n = 4 ⇒ C₄H₄ → \(AgC \equiv C - C \equiv CAg\)

⇒ m_{kết tủa} = 0,05.264.\(\frac{{8,55}}{{12.0,4 + 2.0,45}} = 19,8\)