ancol

L

lop11b

N

niemkieuloveahbu

[TEX]N_{H_2O}=0,45(mol),N_{CO_2}=0,25(mol)\Rightarrow \ ancol\ no,don\ chuc \Rightarrow N_{ancol}=0,2(mol)[/TEX]

Gọi CTTQ: [TEX]R(OH)[/TEX]

[TEX]\Rightarrow M_{ROH}=\frac{7,1}{0,2}=35,5\Rightarrow R=18,5\Rightarrow \overline{n}=1,17\\ \Rightarrow 2\ ancol\ can\ tim: \left{CH_3OH\\C_2H_5OH\right.[/TEX]

[TEX]PT: ROH+CuO\to \ hop\ chat\ cabonyl+ Cu+H_2O[/TEX]

Đều là tỉ lệ 1:1, ở đây hc cabonyl là anđehit

[TEX]\Rightarrow N_{CuO}=N_{ROH}=0,2 \Rightarrow m_{CuO}=16(g)[/TEX]
 
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