a, $ Mg + 2CH_3COOH \rightarrow (CH_3COO)_2Mg + H_2 \\ n_{H_2} = \dfrac{V}{22,4} = \dfrac{6,72}{22,4} = 0,3 (mol) \\ n_{CH_3COOH} = 2.n_{H_2} = 2.0,3 = 0,6 (mol) \\ m_{CH_3COOH} = n . M = 0,6 . 60 = 36 (g) $$\\$ b, $ C\%_{CH_3COOH} = \dfrac{m_{ct}}{m_{dd}}.100\% = \dfrac{36}{250}.100 = 14,4 \% \\ n_{Mg} = n_{H_2} = 0,3 (mol) $$\\$ c, $ m_{Mg} = n . M = 0,3 . 24 = 7,2 (g) $