\(n_{CH_3COOH}=\dfrac{100.12\%}{60}=0,2\left(mol\right)\)
PTHH: CH3COOH + NaHCO3 --> CH3COONa + CO2 + H2O
0,2------>0,2-------------->0,2------->0,2
=> \(m_{dd.NaHCO_3.8,4\%}=\dfrac{0,2.84.100}{8,4}=200\left(g\right)\)
mdd sau pư = 100 + 200 - 0,2.44 = 291,2 (g)
\(m_{CH_3COONa}=0,2.82=16,4\left(g\right)\)
=> \(C\%_{CH_3COONa}=\dfrac{16,4}{291,2}.100\%=5,632\%\)