mCH3COOH = 12g
nCH3COOH = 0.2 mol
CH3COOH + NaHCO3 --> CH3COONa + CO2 + H2O
0.2____________0.2___________0.2_______0.2
mNaHCO3 = 16.8 g
mdd NaHCO3 = 200 g
mCH3COONa = 16.4 g
mCO2 = 8.8 g
mdd sau phản ứng = 100 + 200 - 8.8 = 291.2 g
C%CH3COONa = 16.4/291.2*100% = 5.63%
PTHH.
CH3COOH + NaHCO3 -> CH3COONa + CO2 + H2O
0,2....................0,2..................0,2...............0,2.........0,2 (mol)
Theo bài:
mCH3COOH = \(\frac{100.12}{100}\)= 12 g
=> nCH3COOH = 12/60 = 0,2
Theo pthh và bài có:
nNaHCO3 = nCH3COOH = 0,2 mol
=> mNaHCO3 = 0,2 . 84 = 16,8 g
mddNaHCO3=\(\frac{16,8.100}{8,4}=200\left(g\right)\)
+nCH3COONa = nCH3COOH = 0,2 mol
=> mCH3COONa = 0,2.82 = 16,4 (g)
m dd sau pư = mddCH3COOH + mddNaHCO3− mCO2
= 100 + 200 - 0,2.44= 291,2 (g)
=> C%dd CH3COONa = \(\frac{16,4.100}{291,2}=5,63\left(\%\right)\)
vậy....
mc.t=mCH3COONamc.t== 0,2.82 = 16,4 (gam).
=> mdd=mddCH3COOH+mddNaHCO3−mCO2mdd =mddCH3COOH+mddNaHCO3−mCO2
= 100 + 200 – (0,2.44) = 291,2 (gam).
=> C% (CH3 – COONa) =\(\frac{16,4}{291,2}\) \(\times\)100%= 5,63%.
