Đáp án:

$a\big)C_{M\,CH_3COOH}=0,3M$

$b\big)V_{H_2}=1680(ml)$ 

Giải thích các bước giải:

$a\big)$

Đổi $500ml=0,5l$

$m_{NaOH}=30.20\%=6(g)$

$\to n_{NaOH}=\frac{6}{40}=0,15(mol)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo PT: $n_{CH_3COOH}=n_{NaOH}=0,15(mol)$

$\to C_{M\,CH_3COOH}=\frac{0,15}{0,5}=0,3M$

$b\big)$

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

Theo PT: $n_{H_2}=\frac{1}{2}n_{CH_3COOH}=0,075(mol)$

$\to V_{H_2(đktc)}=0,075.22,4=1,68(l)=1680(ml)$

Đáp án + Giải thích các bước giải:

`a)`

`n_{NaOH}={30.20\%}/{40}=0,15(mol)`

`CH_3COOH+NaOH->CH_3COONa+H_2O`

`n_{CH_3COOH}=n_{NaOH}=0,15(mol)`

`C_{M_{CH_3COOH}}={0,15}/{{500}/{1000}}=0,3M`

`b)`

`Zn+2CH_3COOH->(CH_3COO)_2Zn+H_2`

`n_{H_2}=1/2n_{CH_3COOH}=0,075(mol)`

`V_{H_2}=0,075.22,4=1,68(l)=1680(ml)`