\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+\dfrac{1}{2}H_2\)
0,2 <------------0,2 <------------------------------ 0,1
\(m_{Mg}=0,2.24=4,8\left(g\right)\)
\(m_{ddCH_3COOH}=\dfrac{0,2.60.100}{15}=80\left(g\right)\)