\(n_{MgCO_3}=\dfrac{12.6}{84}=0.15\left(mol\right)\)
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)
\(0.15..........0.3............................................0.15\)
\(V_{CO_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{CH_3COOH}=0.3\cdot60=18\left(g\right)\)
\(C\%_{CH_3COOH}=\dfrac{18}{200}\cdot100\%=9\%\)
Pthh MgCO3+2CH3COOH--->(CH3COO)2Mg+CO2+H2O
Ta có nMgCO3=0,15 mol
Theo pthh thì nCH3COOH=0,3 mol
=>mCH3COOH=17,7 g
=>C%CH3COOH=8,85%