Đáp án:

$a\big)V_{H_2}=4,48(l)$

$b\big)m_{(CH_3COO)_2Zn}=36,6(g)$

$c\big)C_{M\,(CH_3COO)_2Zn}=1M$ 

Giải thích các bước giải:

$a\big)$

Đổi $200ml=0,2l$

$\to n_{CH_3COOH}=0,2.2=0,4(mol)$

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

Theo PT: $n_{H_2}=\frac{1}{2}n_{CH_3COOH}=0,2(mol)$

$\to V_{H_2(đktc)}=0,2.22,4=4,48(l)$

$b\big)$

Theo PT: $n_{(CH_3COO)_2Zn}=n_{H_2}=0,2(mol)$

$\to  m_{(CH_3COO)_2Zn}=0,2.183=36,6(g)$

$c\big)$

$C_{M\,(CH_3COO)_2Zn}=\frac{0,2}{0,2}=1M$

Đáp án + Giải thích các bước giải:

`a)`

`n_{CH_3COOH}={200}/{1000}.2=0,4(mol)`

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

`n_{H_2}=1/2n_{CH_3COOH}=0,2(mol)`

`V_{H_2}=0,2.22,4=4,48(l)`

` b)`

`n_{(CH_3COO)_2Zn}=n_{H_2}=0,2(mol)`

`m_{(CH_3COO)_2Zn}=0,2.183=36,6(g)`

`c)`

`C_{M_{(CH_3COO)_2Zn}}={0,2}/{{200}/{1000}}=1M`