Question

Answer 1

a) Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

b) \(m_{Na_2CO_3}=\dfrac{250.8,48}{100}=21,2\left(g\right)\Rightarrow n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

                  0,2--------->0,4--------------->0,4------->0,2

V = 0,2.22,4 = 4,48 (l)

\(m_{CH_3COOH}=0,4.60=24\left(g\right)\Rightarrow m=\dfrac{24.100}{8}=300\left(g\right)\)

c) m_{dd sau pư} = 250 + 300 - 0,2.44 = 541,2 (g)

=> \(C\%_{dd.CH_3COONa}=\dfrac{0,4.82}{541,2}.100\%=6,06\%\)