CH3COOH + NaHCO3 \(\rightarrow\) CH3COONa + H2O + CO2\(\uparrow\) (1)
a) Ta có: \(m_{NaHCO_3}=\dfrac{20\cdot500}{100}=100\left(g\right)\)
\(\Rightarrow n_{NaHCO_3}=\dfrac{100}{84}\approx1,2\left(mol\right)\)
Theo phương trình (1): \(n_{CH_3COOH}=n_{NaHCO_3}=1,2\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=1,2\cdot60=72\left(g\right)\)
Do đó: \(C\%_{CH_3COOH}=\dfrac{72}{300}\cdot100\%=24\%\)