CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O
mCH3COOH = 100x12/100 = 12 (g)
==> nCH3COOH = m/M = 12/60 = 0.2 (mol)
Theo pt: => nNaHCO3 = 0.2 (mol)
==> mNaHCO3 = n.M = 0.2x84 =16.8 (g)
==> mdd NaHCO3 = 16.8x100/8.4 = 200 (g)
Ta có: nCH3COONa = 0.2 (mol)
==> mCH3COONa = n.M = 0.2 x 82 = 16.4 (g)
mdd sau pứ = 200 + 100 - 0.2 x 44 =291.2 (g)
C% = 16.4 x 100/ 291.2 = 5.63%