PTHH:

$CH_3-CH_2-OH$ + $CuO$ $\overset{t^o}{\rightarrow}$ $CH_3CHO$ + $Cu$ + $H_2O$

$CH_3CHO$ + $2AgNO_3$ + $NH_3$ + $H_2O$ $\overset{}{\rightarrow}$ $CH_3COONH_4$ + $2Ag$ +$NH_4NO_3$

 số mol etanol=$\frac{3}{46}=3/46mol$

=> số mol axetandehit giả định = $3/46mol$

ta có:số mol bạc = 2 lần số mol axetandehit

=>số mol axetandehit thực tế = $\frac{8,1}{108}:2=0,0375mol$

H=$\frac{^n thực}{^n GD}$=$\frac{0,0375}{3/46}.100$% = 57,5%

$C_2H_5OH+CuO\buildrel{{t^o}}\over\to CH_3CHO+Cu+H_2O$

$CH_3CHO\buildrel{{AgNO_3/NH_3}}\over\longrightarrow 2Ag$

$n_{Ag\downarrow}=\dfrac{8,1}{108}=0,075(mol)$

$\to n_{CH_3CHO}=\dfrac{n_{Ag}}{2}=0,0375(mol)$

$\to n_{C_2H_5OH\text{pứ}}=n_{CH_3CHO}=0,0375(mol)$

$\to H=\dfrac{0,0375.46.100}{3}=57,5\%$