\(a,m_{CH_3COOH}=\dfrac{12.200}{100}=24\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{24}{60}=0,4\left(mol\right)\)
PTHH: CH3COOH + NaHCO3 ---> CH3COONa + CO2 + H2O
0,4---------->0,4--------------------------------->0,4
\(\rightarrow C\%_{NaHCO_3}=\dfrac{0,4.106}{400}.100\%=10,6\%\)
b, VCO2 = 0,4.22,4 = 8,96 (l)