a, \(n_{H_2}=n_C=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 2CH3COOH + Mg ---> (CH3COO)2Mg + H2
0,2<-------0,1<---------0,1<--------------0,1
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{MgO}=8,4-2,4=6\left(g\right)\end{matrix}\right.\\ n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{8,4}.100\%=28,57\%\\\%m_{MgO}=100\%-28,57\%=71,53\%\end{matrix}\right.\)
b, PTHH: MgO + 2CH3COOH ---> (CH3COO)2Mg + H2O
0,15---->0,3----------------->0,15
=> \(\left\{{}\begin{matrix}m_{ddB}=8,4+\dfrac{\left(0,2+0,3\right).60}{9\%}-0,1.2=341,53\left(g\right)\\m_{\left(CH_3COO\right)_2Mg}=\left(0,15+0,1\right).142=35,5\left(g\right)\end{matrix}\right.\\ \Rightarrow C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{35,5}{341,53}.100\%=10,4\%\)
