Question

Answer 1

a)

\(n_{CH_3COOH}=\dfrac{200.12\%}{60}=0,4\left(mol\right)\)

PTHH: CH₃COOH + NaHCO₃ --> CH₃COONa + CO₂ + H₂O

                  0,4------->0,4----------->0,4---------->0,4

=> \(m_{NaHCO_3}=0,4.84=33,6\left(g\right)\)

=> \(m=\dfrac{33,6.100}{8,4}=400\left(g\right)\)

b) 

\(m_{CH_3COONa}=0,4.82=32,8\left(g\right)\)

m_{dd sau pư} = 200 + 400 - 0,4.44 = 582,4 (g)

\(C\%_{CH_3COONa}=\dfrac{32,8}{582,4}.100\%=5,632\%\)

c) m_CO2 = 0,4.44 = 17,6 (g)