a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Ta có: \(n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=n_{C_2H_4Br_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{4,48}.100\%=50\%\\\%V_{CH_4}=50\%\end{matrix}\right.\)

b, \(C\%_{Br_2}=\dfrac{0,1.160}{100}.100\%=16\%\)

- Trích mẫu thử.

- Nhỏ vài giọt từng mẫu thử vào giấy quỳ tím.

+ Quỳ hóa xanh: NaOH

+ Quỳ hóa đỏ: HCl

+ Quỳ không đổi màu: C₆H₁₂O₆, C₂H₅OH (1)

- Cho mẫu thử nhóm (1) pư với dd AgNO₃/NH₃

+ Có tủa sáng bạc: C₆H₁₂O₆

PT: \(CH_2OH\left[CHOH\right]_4CHO+2AgNO_3+3NH_3+H_2O\underrightarrow{^{t^o}}CH_2OH\left[CHOH\right]_4COONH_4+2NH_4NO_3+2Ag\)

+ Không hiện tượng: C₂H₅OH

- Dán nhãn.

a, \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

 \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

0,05__________0,1_____________________0,05 (mol)

\(\Rightarrow m_{Mg}=0,05.24=1,2\left(g\right)\)

b, \(n_{C_2H_5OH}=\dfrac{12,9-0,1.60}{46}=0,15\left(mol\right)\)

PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (đk: t^o, H₂SO₄)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\), ta được C₂H₅OH dư.

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,1.88=8,8\left(g\right)\)

\(\Rightarrow H\%=\dfrac{5,72}{8,8}.100\%=65\%\)

a, \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

b, Ta có: \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=1.60=60\left(g\right)\)

\(\Rightarrow m_{C_2H_5OH}=90-60=30\left(g\right)\)

c, \(n_{C_2H_5OH}=\dfrac{30}{46}=\dfrac{15}{23}\left(mol\right)\)

\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (đk: t^o, H₂SO₄ đặc)

Xét tỉ lệ: \(\dfrac{1}{1}>\dfrac{\dfrac{15}{23}}{1}\), ta được CH₃COOH dư.

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{C_2H_5OH}=\dfrac{15}{23}\left(mol\right)\)

Mà: H = 90%

\(\Rightarrow n_{CH_3COOC_2H_5\left(TT\right)}=\dfrac{15}{23}.90\%=\dfrac{27}{46}\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(TT\right)}=\dfrac{27}{46}.88=\dfrac{1188}{23}\left(g\right)\)

a, \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

b, \(n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{CO_2}=2n_{C_2H_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,4.22,4=8,96\left(l\right)\)

c, \(C_2H_4+H_2O\underrightarrow{xt}C_2H_5OH\)

\(n_{C_2H_5OH\left(LT\right)}=n_{C_2H_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(LT\right)}=0,2.46=9,2\left(g\right)\)

Mà: H = 80%

⇒ m_{C2H5OH (TT)} = 9,2.80% = 7,36 (g)

d, n_C2H5OH = 0,2.80% = 0,16 (mol)

PT: \(C_2H_5OH+O_2\underrightarrow{^{t^o,xt}}CH_3COOH+H_2O\)

_____0,16______________0,16 (mol)

\(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)

_______0,16___________________0,16_____0,08 (mol)

m _{dd sau pư} = 7,36 + 200 - 0,08.44 = 203,84 (g)

\(\Rightarrow C\%_{CH_3COOK}=\dfrac{0,16.98}{203,84}.100\%\approx7,7\%\)