Câu 14:

Ancol X có không quá 3C, pư với Cu(OH)₂ tạo dd xanh lam.

→ CH₂(OH)-CH₂OH

CH₂(OH)-CH₂(OH)-CH₂OH

CH₃-CH₂(OH)-CH₂OH

Đáp án: C

Câu 15:

\(n_{C_6H_5OH}=\dfrac{1,41}{94}=0,015\left(mol\right)\)

PT: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr\)

Theo PT: \(n_{Br_2}=3n_{C_6H_5OH}=0,045\left(mol\right)\)

Đáp án: B

Câu 9:

\(n_{CH_3CHO}=\dfrac{5,28}{44}=0,12\left(mol\right)\)

PT: \(CH_3CHO+H_2\underrightarrow{t^o,Ni}CH_3CH_2OH\)

Theo PT: \(n_{H_2}=n_{CH_3CHO}=0,12\left(mol\right)\Rightarrow V_{H_2}=0,12.22,4=2,688\left(l\right)\)

Đáp án: A

Câu 12:

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PT: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

Theo PT: \(n_{C_2H_5OH}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH}=0,3.46=13,8\left(g\right)\)

Đáp án: D

PTHH :

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

 trc p/u :                                     0,1        0,15

p/u :                   0,05    0,15    0,1           0,1 

sau p/u:           0,05       0,15      0           0,05 

-----> H2O dư 

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{H_2O}=\dfrac{2,7}{18}=0,15\left(mol\right)\)

\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

Ta có: \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{H_2O}=\dfrac{2,7}{18}=0,15\left(mol\right)\)

BTNT O, có: 2n_O2 = 2n_CO2 + n_H2O

⇒ n_O2 = 0,175 (mol)

\(\Rightarrow V_{O_2}=0,175.22,4=3,92\left(l\right)\)

\(CTTQ:C_aH_{2a+1}OH\left(a:nguyên,dương\right)\\ C_aH_{2a+1}OH+Na\rightarrow C_aH_{2a+1}ONa+\dfrac{1}{2}H_2\\ n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ n_{C_aH_{2a+1}OH}=2.0,05=0,1\left(mol\right)\\ M_{C_aH_{2a+1}OH}=\dfrac{3,2}{0,1}=32\left(\dfrac{g}{mol}\right)\\ M_{C_aH_{2a+1}}=32-17=15\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow14a+1=15\\ \Leftrightarrow a=1\\ Vậy.CTPT.X:CH_4O\)