$ n_{Mg} = \dfrac{m}{M} = \dfrac{4,8}{24} = 0,2 (mol) \\ m_{CH_3COOH} = 360 . 5\% = 18 (g) \\ n_{CH_3COOH} = \dfrac{m}{M} = \dfrac{18}{60} = 0,3 (mol) \\ 2CH_3COOH + Mg \rightarrow (CH_3COOH)_2Mg + H_2 $
Xét tỷ lệ giữa $CH_3COOH$ và $Mg$ : $\dfrac{0,3}{2} < \dfrac{0,2}{1}$ ( $n$ $Mg$ dư tính theo $n$ $CH_3COOH$ )
$m_{H_2} = n . M = 0,15 . 2 = 0,3 (g) \\ m_{Mgpứ} = n . M = 0,15 . 24 = 3,6 (g) \\ m_{ddspứ} = m_{Mgpứ} + m_{CH_3COOH} - m_{H_2} = 3,6 + 360 - 0,3 = 363,3 (g) \\ m_{ct(CH_3COO)_2Mg} = n .M = 0,15 . 142 = 21,3 (g) \\ \rightarrow C\%_{(CH_3COO)_2Mg} = \dfrac{m_{ct}}{m_{dd}}.100\% = \dfrac{21,3}{363,3}.100 = 5,86 \% $
