\(n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
PTHH: CH3COOH + MHCO3 --> CH3COOM + CO2 + H2O
0,2------->0,2--------------------->0,2
=> VCO2 = 0,2.22,4 = 4,48 (l)
\(m_{MHCO_3}=\dfrac{200.10}{100}=20\left(g\right)\)
=> \(M_{MHCO_3}=\dfrac{20}{0,2}=100\left(g/mol\right)\)
=> MM = 39 (g/mol)
=> M là K