<p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">a) Các phương trình phản ứng xảy ra:</span></p><p class="MsoNormalCxSpMiddle" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">C<sub>2</sub>H<sub>5</sub>OH + Na → C<sub>2</sub>H<sub>5</sub>ONa +<script type="math/tex">\frac{1}{2}</script>H<sub>2</sub></span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">CH<sub>3</sub>COOH + Na → CH<sub>3</sub>COONa +<script type="math/tex">\frac{1}{2}</script>H<sub>2</sub></span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">C<sub>6</sub>H<sub>5</sub>OH + Na → C<sub>6</sub>H<sub>5</sub>ONa + <script type="math/tex">\frac{1}{2}</script></span><span style=" ">H<sub>2</sub></span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">C<sub>6</sub>H<sub>5</sub>OH + 3Br<sub>2</sub> → C<sub>6</sub>H<sub>2</sub>OH(Br)<sub>3</sub> + 3HBr</span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">b) Số mol H<sub>2</sub> là:<script type="math/tex">{n}_{{H}_{2}}=\frac{4,48}{22,4}</script>= 0,2 mol</span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">Gọi số mol </span><span style="  mso-fareast- mso-bidi-font-weight: bold;">chất ancol etylic, axit axetic, phenol lần lượt là x, y, z (mol), (x, y, z >0)</span></p><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" "><script type="math/tex">\begin{array}{l}{C}_{2}{H}_{5}OH\text{ }+\text{ }Na\text{ }\to \text{ }{C}_{2}{H}_{5}ONa\text{ }+\text{ }\frac{1}{2}\text{ }{H}_{2}\ x\text{                                                              }\frac{x}{2}\text{    mol}\end{array}</script></span></p><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" "><script type="math/tex">\begin{array}{l}C{H}_{3}COOH\text{ }+\text{ }Na\text{ }\to \text{ }C{H}_{3}COONa\text{ }+\text{ }\frac{1}{2}\text{ }{H}_{2}\ y\text{                                                                   }\frac{y}{2}\text{      mol}\end{array}</script></span></p><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" "><script type="math/tex">\begin{array}{l}{C}_{6}{H}_{5}OH\text{ }+\text{ }Na\text{ }\to \text{ }{C}_{6}{H}_{5}ONa\text{ }+\frac{1}{2}{H}_{2}\ z\text{}\frac{z}{2}\text{}mol\end{array}</script></span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto; tab-stops: .5in 1.0in 1.5in 2.0in 2.5in 3.0in center 274.5pt left dotted 7.0in;"><span style=" ">Số mol kết tủa </span><span style="  mso-fareast- mso-bidi-font-weight: bold;">trắng 2,4,6-tribromphenol là: </span><span style=" ">= 0,2 mol </span></p><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto; tab-stops: .5in 1.0in 1.5in 2.0in 2.5in 3.0in center 274.5pt left dotted 7.0in;"><span style="  mso-ansi-language: PT-BR;"><script type="math/tex">\begin{array}{l}{C}_{6}{H}_{5}OH\text{ }+\text{ }3B{r}_{2}\to \text{ }{C}_{6}{H}_{2}\left(OH\right){\left(Br\right)}_{3}+\text{ }3HBr\ z\text{                                            z                                     mol}\end{array}</script></span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">Suy ra z = 0,2 mol</span></p><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" "><script type="math/tex">\left\{\begin{array}{l}46x+60y+94z=29,4\ \frac{1}{2}x+\frac{1}{2}y+\frac{1}{2}z=0,2\ 331z=66,2\end{array}\right.</script></span></p><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">Suy ra <script type="math/tex">\left\{\begin{array}{l}x=0,1\ y=0,1\ z=0,2\end{array}\right.</script></span></p><div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">Vậy % ancol etylic = 25%, % axit axetic = 25% và %phenol = 50%</span></p></div></div></div></div></div></div></div></div>

<div><p class="MsoNormalCxSpFirst" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">(1) CH<sub>3</sub>-CH<sub>2</sub>-CH<sub>3</sub><script type="math/tex">\stackrel{cracking}{\to}</script>CH<sub>2</sub>=CH<sub>2</sub> + CH<sub>4</sub></span></p><p class="MsoNormalCxSpMiddle" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">(2) CH<sub>2</sub>=CH<sub>2</sub> + H<sub>2</sub>O <script type="math/tex">\stackrel{{H}^{+}}{\to}</script></span><span style="  mso-ansi-language: IT;">CH<sub>3</sub>-CH<sub>2</sub>-OH </span></p><p class="MsoNormalCxSpMiddle" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">(3) CH<sub>3</sub>-CH<sub>2</sub>-OH + CuO <script type="math/tex">\stackrel{{t}^{o}}{\to}</script></span><span style="  mso-ansi-language: IT;">CH<sub>3</sub>-CHO + Cu + H<sub>2</sub>O </span></p><p class="MsoNormalCxSpMiddle" style="margin-top: 0in; mso-add-space: auto;"><span style=" ">(4) CH<sub>3</sub>-CHO + 2AgNO<sub>3 </sub>+ 2NH<sub>3</sub> → CH<sub>3</sub>-COONH<sub>4</sub> + 2Ag + NH<sub>4</sub>NO<sub>3 </sub></span></p></div>

Bộ Đề kiểm tra Hóa 11 học kì 2 có đáp án (đề 18)

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Giải bài tập tự luận: Bộ Đề kiểm tra Hóa 11 học kì 2 có đáp án (đề 18), thời gian làm 12 phút/2 câu. Biên soạn bởi gia sư Lê Phương Thảo.

Bộ Đề kiểm tra Hóa 11 học kì 2 có đáp án (đề 18)

12-05-2026

00:12:00