$n_{Mg}=\dfrac{8,4}{24}=0,35(mol)$
$2CH_3COOH+Mg\to (CH_3COO)_2Mg+H_2$
$\to n_{CH_3COOH}=2n_{Mg}=0,35.2=0,7(mol)$
$\to m_{\rm dd CH_3COOH}=0,7.60:40\%=105g$
$n_{(CH_3COO)_2Mg}=n_{H_2}=n_{Mg}=0,35(mol)$
$\to m_{dd\rm spứ}=8,4+105-0,35.2=112,7g$
$\to C\%_{(CH_3COO)_2Mg}=\dfrac{0,35.142.100}{112,7}=44,1\%$