Hỗn hợp X gồm \(\mathrm{C}_{2} \mathrm{H}_{4}, \mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{C}_{3} \mathrm{H}_{8}, \mathrm{C}_{4} \mathrm{H}_{10} .\) Lấy \(6,32\,\text{gam}\,\text{X}\) cho qua bình đựng dung dịch nước \(\mathrm{Br}_{2}\) (đủ) thấy có \(0,12\,~\text{mol}\,\,\text{B}{{\text{r}}_{2}}\) tham gia phản ứng. Mặt khác, đốt cháy hết 2,24 lít X (đktc) cần dùng V lít khí \(O_{2}\) (đktc), sau phản ứng thu được 9,68 gam \(\mathrm{CO}_{2}\). Biết các phản ứng hoàn toàn. Giá trị của V là

Đáp án C.

\(0,1\,\,~\text{mol}\,\,\,\text{X}:{{\text{C}}_{x}}{{\text{H}}_{y}}\to \text{C}{{\text{O}}_{2}}:0,22~\,\,\text{mol}\)

\(n_{\mathrm{x}}=\frac{n_{\mathrm{CO}_{2}}-n_{\mathrm{H}_{2} \mathrm{O}}}{k-1} \rightarrow n_{\mathrm{CO}_{2}}-n_{\mathrm{H}_{2} \mathrm{O}}=n_{\mathrm{x}} k-n_{\mathrm{x}}\)

\(\to 0,22-{{n}_{{{\text{H}}_{2}}\text{O}}}={{n}_{\partial }}-0,1\,(1)\)

\(m_{\mathrm{x}}=m_{\mathrm{C}}+m_{\mathrm{H}}=12 n_{\mathrm{CO}_{2}}+2 n_{\mathrm{H}_{2} \mathrm{O}}=12.0,22+n_{\mathrm{H}_{2} \mathrm{O}}=\left(2,64+2 n_{\mathrm{H}_{2} \mathrm{O}}\right)(\mathrm{g})\)

\(\left\{ \begin{array}{*{35}{l}} {{m}_{\text{X}}}:\left( 2,64+2{{n}_{{{\text{H}}_{2}}\text{O}}} \right)\to n={{n}_{\text{B}{{\tau }_{2}}}} \\ {{m}_{\text{X}}}:6,32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to {{n}_{{{\text{B}}_{2}}}}=0,12 \\ \end{array}\to \frac{2,64+2{{n}_{{{\text{H}}_{2}}\text{O}}}}{6,32}=\frac{{{n}_{\partial }}}{0,12}(2) \right.\)

\(\to \left\{ \begin{array}{*{35}{l}} {{n}_{{{\text{H}}_{2}}\text{O}}}=0,26 \\ {{n}_{\partial }}=0,06 \\ \end{array} \right.\)

BT \(\text{O}:{{n}_{{{\text{O}}_{2}}}}=\frac{2{{n}_{\text{C}{{\text{O}}_{2}}}}+{{n}_{{{\text{H}}_{2}}\text{O}}}}{2}=0,35~\text{mol}\)

\(\rightarrow V=7,84\) (l).