Đáp án + Giải thích các bước giải:

$n_{CH_3COOH}=\frac{200.22,5\%}{60}=0,75(mol)$

$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$

Theo PT: $n_{Zn}=n_{(CH_3COO)_2Zn}=n_{H_2}=0,375(mol)$

$\to m_{Zn}=0,375.65=24,375(g)$

$\to C\%_{(CH_3COO)_2Zn}=\frac{0,375.183}{24,375+200-0,375.2}.100\%\approx 30,69\%$

Đáp án:

`↓` 

Giải thích các bước giải:

`m_{CH_3COOH}=``\frac{C%.m_{dd}}{100%}=\frac{22,5%.200}{100%}=45(g)`

`n_{CH_3COOH}=``\frac{m}{M}=\frac{45}{60}=0,75(mol)`

`PTHH`             `Zn``+``2CH_3COOH``→``(CH_3COO)_2Zn``+``H_2↑`

                     `0,375`          `0,75`                     `0,375`            `0,375`          `mol`

`→n_{Zn}=``\frac{1}{2}n_{CH_3COOH}=\frac{0,75.1}{2}=0,375(mol)`

`⇒m_{Zn}=n.M=0,375.65=24,375(g)`

`→n_{H_2}=``\frac{1}{2}n_{CH_3COOH}=\frac{0,75.1}{2}=0,375(mol)`

`m_{H_2}=n.M=0,375.2=0,75(g)`

`m_{sau}=m_{Zn}+m_{dd_{CH_3COOH}}-m_{H_2}=24,375+200-0,75=223,625(g)`

`→n_{(CH_3COO)_2Zn}=``\frac{1}{2}n_{CH_3COOH}=\frac{0,75.1}{2}=0,375(mol)`

`⇒C%_{(CH_3COO)_2Zn}=``\frac{m_{(CH_3COO)_2Zn}.100%}{m_{sau}}=\frac{0,375.183.100%}{223,625}≈30,69%`