`2CH_3COOH+Mg->(CH_3COO)_2Mg+H_2uparrow`

`n_((CH_3COO)_2Mg)=(2,13)/(142)=0,015mol`

`=>{(n_(CH_3COOH)=\text(0,015.2=0,03)mol),(n_(H_2)=\text(0,015)mol):}`

`a.`

`C_M(CH_3COOH)=(0,03)/(200.10^-3)=0,15M`

`b.`

`V_(H_2)(đktc)=0,015.22,4=0,336l`

`c.`

`CH_3COOH+NaOH->CH_3COONa+H_2O`

`n_(ddCH_3COOH)=0,15.75.10^-3=0,01125mol`

`=>n_(NaOH)=0,01125mol`

`V_(NaOH)=(0,01125)/(0,5)=0,0225l=22,5l`

Phản ứng xảy ra:

$\rm Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2(1)$

Theo đề:

$\rm n_{(CH_3COO)_2Mg} = \dfrac{2,13}{142} = 0,015$ mol

a)Theo PTHH$\rm (1)$:

$\rm n_{CH_3COOH} = 2n_{(CH_3COO)_2Mg} = 2 . 0,015 = 0,03$ mol

Đổi $\rm 200ml = 0,2l$

Nồng độ mol dd $\rm CH_3COOH$ đã dùng:

$\rm C_{M\text{ dd } CH_3COOH} = \dfrac{0,03}{0,2} = 0,15M$

b)Theo PTHH$\rm (1)$:

$\rm n_{H_2} = n_{(CH_3COO)_2Mg} = 0,015$ mol

Thể tích khí $\rm H_2$ thoát ra ở đktc:

$\rm V_{H_2} = 0,015 . 22,4 = 0,336L$

c)

$\rm CH_3COOH + NaOH \to CH_3COONa + H_2O(2)$

Đổi $\rm 75ml = 0,075l$

Trong $\rm 0,2L$ dd $\rm CH_3COOH$ chứa $\rm 0,03$ mol

Vậy trong $\rm 0,075L$ dd $\rm CH_3COOH$ chứa $\rm x(mol)$

$\rm \Rightarrow x = n_{CH_3COOH(75ml)} = \dfrac{0,075 . 0,03}{0,2} = 0,01125$ mol

Theo PTHH$\rm (2):$

$\rm n_{NaOH} = n_{CH_3COOH(75ml)} = 0,01125$ mol

Thể tích dd $\rm NaOH$ cần:

$\rm V_{\text{dd }NaOH} = \dfrac{0,01125}{0,5} = 0,0225L = 22,5ml$