Question

Problema 3* ultima dau coroana

Answer 1

F.g acizi monocarboxilici saturati : CnH2nO2   = 14n+32

3. a) %O = mO/Macizi*100 <=> 53,33 = 32/(14n+32)*100 <=> 53,33(14n+32) = 3200 => n = 2. 
F.M acid -> H3C-COOH - acid acetic

b) Vs = ?
C = 0,4  M
ms = 200g
C = 60%

C  = md/ms*100 => md = C*ms/100 = 60*200/100 = 120g acid

n acid = 120/60 = 2 moli 

H3C-COOH + NaOH => H3C-COONa + H2O

n acid = n NaOH = 2 moli

C = n/Vs => Vs = n/C = 2/0,4 = Calculeaza Vs in L !!

c) Vs = ? 
C = 4M

m = 199,2g CaO impur

nCaO/nCaCO3 = 16:1

56x + 100y = 199,2 => 56*16y + 100y = 199,2 => y = 0,2 moli

x/y = 16/1 => x=16y = 3,2 moli

m CaO = 3,2*56 = 179,2g

m CaCO3 = 0,2*100 = 20g

2 moli................1 mol
2H3C-COOH + CaO ---> (H3C-COO)2Ca + H2O
x moli ...............3,2 moli

x = 6,4 moli acid

2 moli................1 mol........................................1 mol
2H3C-COOH + CaCO3 ---> (H3C-COO)2Ca + CO2 + H2O
y moli................0,2 moli.....................................z moli

y = 0,4  moli acid
z = 0,2 moli CO2

n acid = 0,4+6,4 = 6,8 moli

C = n/Vs => Vs = n/C = 6,8/4 = 1,7L sol, H3C-COOH

d) n CO2 = 0,2 moli

pV = nRT => V = nRT/p = (0,2*0,082*300)/3 = 1,64 L CO2

Answer 2

a) R-COOH    CxH2xO2
32·100/M = 53,33    M = 60 g/mol      14x + 32 = 60  x = 2    CH3-COOH
b) CH3COOH + NaOH = CH3COONa + H2O
nNaOH = nCH3COOH
0,4Vb = 200x0,6 /100      V = 3 l
c) 2CH3COOH + CaO = (CH3COO)2Ca + H2O
2CH3COOH + CaCO3 = (CH3COO)2Ca + CO2 + H2O
n1 moli CaO
n2 moli  CaCO3       56n1 + 100n2 = 199,2     
14n1 + 25n2 = 49,8
n1 = 16n2       224n2 + 25n2 = 49,8   249n2 = 49,8      n2 = 0,2 moli CaCO3
n1 = 3,2 moli CaO
nCH3COOH = 2n1+ 2n2 = 6,8 moli      Vs = 6,8/4 = 1,7 l
d) nCO2 = n2/2 = 0,1moli
PV = nRT        V = 0,1x0,082x300/3 = 0,82 l