\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,3<------0,6<---------------0,3<------0,3
=> \(C\%_{\left(dd.CH_3COOH\right)}=\dfrac{0,6.60}{250}.100\%=14,4\%\)
\(m_{\left(CH_3COO\right)_2Mg}=0,3.142=42,6\left(g\right)\)