2.
n moli 2 moli
2CH3COOH + Ca --> (CH3COO)2Ca + H2
2 1
=> n = 2x2/1 = 4 moli CH3COOH
stim ca Cm = nr.moli(n)/Vs
=> Vs = n/Cm = 4/0,1 = 40 L sol.
3.
a.
stim ca c% = mdx100/ms
md g 12 g
2CH3COOH + Mg --> (CH3COO)2Mg + H2
2x60 24
=> md = 2x60x12/24 = 60 g CH3COOH
=> c% = 60x100/200 = 30%
b.
Vs.NaOH = 800 mL = 0,8 L, Cm = 0,5 M
stim ca Cm = n/Vs
=> n = CmxVs = 0,8x0,5 = 0,4 moli NaOH
n moli 0,4 moli
CH3COOH + NaOH --> CH3COONa + H2O
1 1
=> n = 1x0,4/1 = 0,4 moli CH3COOH
=> Cm = 0,4/0,2 = 2 mol/L
c.
Vs.NaOH = 250 cm3 = 0,25 L, Cm = 4 M
stim ca Cm = n/Vs
=> n = CmxVs = 4x0,25 = 1 mol NaOH
md g 1 moli
CH3COOH + NaOH --> CH3COONa + H2O
60 1
=> md = 60x1/1 = 60 g CH3COOH
c% = mdx100/ms
= 60x100/200 = 30%
