Đáp án + Giải thích các bước giải:

 `1)`

$C_2H_4 + 3O_2 \xrightarrow{t^o}2CO_2 + 2H_2O$

`n_(CO_2)=2n_(C_2H_4)=2.0,3=0,6(mol)`

`2)`

`-` Trong `10` ml dung dịch rượu có:

`V_(C_2H_5OH)={46.10}{100}=4,6(ml)`

`3)`

`-` PTHH:

$CH_3CHO + 2AgNO_3 + 3NH_3 + H_2O \longrightarrow CH_3COONH_4 + 2Ag\downarrow  + 2NH_4NO_3$

`->` Có: `n_(Ag)=n_(CH_3CHO)=2.0,2=0,4(mol)`

`->` `m_(Ag)=0,4.108=43,2(g)`

`4)`

`-` PTHH:

$CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O$

`+)`$n_{CH_3COOH} = \frac{m_{CH_3COOH}}{M_{CH_3COOH}} = \frac{60}{60} = 1 (mol)$

`+)` $n_{CH_3COOC_2H_5} = \frac{m_{CH_3COOC_2H_5}}{M_{CH_3COOC_2H_5}} = \frac{66}{88} = 0,75 (mol)$

`-` Có:

`H%={n_(CH_3COOC_2H_5)}/{n_(CH_3COOH)}.100%`

`={0,75}/{1}.100%=75%`