$\rm PTHH: Fe +2CH_3COOH \xrightarrow{} (CH_3COO)_2Fe +H_2$ 

`n_{CH_3COOH}=\frac{300.3%}{60}=0,15(mol)`

theo pthh 

`n_{H_2}=n_{Fe}=n_{(CH_3COO)_2Fe}=\frac{1}{2} n_{CH_3COOH}=0,075(mol)`

a)

`m_{Fe}=a=0,075.56 =4,2(g)`

`V=V_{H_2}=0,075.24,79=1,85925(l)`

b)

`m_{dd-sp ứ}=300+4,2-0,075.2=304,05(g)`

$m_{(CH_3COO)_2Fe}=0,075.174=13,05(g)$ 

`->C%_{(CH_3COO)_2Fe}=\frac{13,05}{304,05}.100%=4,3%`

Đáp án:

Giải thích các bước giải:

 `m_(CH_3 COOH) = (300 xx 3% )/(100 %) =9(g)`

`n_(CH_3 COOH) = 9/60=0,15 (mol)`

`PTHH: 2CH_3 COOH + Fe -> (CH_3 COO)_2 Fe + H_2 ↑`

               `0,15            -> 0,075   -> 0,075 -> 0,075`

`a)` `m_Fe = 0,075 xx 56 =4,2 (g)`

`V_(H_2 ) =22,4 xx 0,075 =1,68(l)`

`m_(H_2) = 2xx 0,075 =0,15(g)`

`b) ``m_((CH_3 COO)_2 Fe) =0,075 xx 174 =13,05(g)`

`m_(dd(CH_3 COO)_2 Fe) =m_(ddCH_3 COOH) + m_Fe -m(H_2) = 300+4,2 -0,15=304,05(g)`

`C%_((CH_3COO)_2 Fe)=13,05/304,05 xx 100%=4,3%`