Giải thích các bước giải:

$a)$

$n_K=\dfrac{15,6}{39}=0,4(mol)$

$K+CH_3COOH\to CH_3COOK+\dfrac{1}{2}H_2$

$n_{CH_3COOK}=n_K=0,4(mol)$

$m_{CH_3COOK}=0,4×98=39,2(g)$ 

$b)$

$n_{CH_3COOH}=n_K=0,4(mol)$

$m_{dd\ CH_3COOH}=\dfrac{0,4×60}{10\%}=240(g)$

$c)$

$n_{H_2}=\dfrac{1}{2}n_K=0,2(mol)$

$V_{H_2}=0,2×22,4=4,48(l)$

$d)$

$m_{dd\ sau}=15,6+240-0,2×2=255,2(g)$

$C\%_{CH_3COOK}=\dfrac{39,2}{255,2}×100\%=15,361\%$

Đáp án:

`↓` 

Giải thích các bước giải:

`a.` `n_K=``\frac{m}{M}=\frac{15,6}{39}=0,4(mol)`

`PTHH`            `2K``+``2CH_3COOH``→``2CH_3COOK``+``H_2↑`

                       `0,4`            `0,4`                     `0,4`             `0,2`             `mol`

`→n_{CH_3COOK}=n_K=0,4(mol)`

`→m_{CH_3COOK}=n.M=0,4.98=39,2(g)`

`b.`Theo pt, ta có: `n_{CH_3COOH}=n_K=0,4(mol)`

`→m_{CH_3COOH}=n.M=0,4.60=24(g)`

`c.`Theo pt, ta có: `n_{H_2}=``\frac{1}{2}n_K=\frac{0,4.1}{2}=0,2(mol)`

`→V_{H_2(đktc)}=n.22,4=0,2.22,4=4,48(l)`

`d.``m_{H_2}=n.M=0,2.2=0,4(g)`

`m_{dd_{CH_3COOH}}=``\frac{m.100%}{C%}=\frac{24.100%}{10%}=240(g)`

`m_{sau}=m_K+m_{dd_{CH_3COOH}}-m_{H_2}=15,6+240-0,4=255,2(g)`

`C%_{CH_3COOK}=``\frac{m.100%}{m_{sau}}=\frac{39,2.100%}{255,2}≈15,36%`