Question

Un acid monocarboxilic saturat aciclic are un continut in oxigen de 43,24%. Calculati volumul solutiei de acid,de concentratie 2M,care reactioneaza cu 193 g oxid de calciu carbonatat,stiind ca raportul molar CaO:CaCO3=12:1

Answer 1

CnH2nO2 => 14n+32

%O=mO/M*100 => 43,24=32/14n+32*100 => n=3

C3H6O2 -> acid propanoic

56x+100y=193
x/y=12/1 => x=12y=>x=3 moli

56*12y+100y=193 => y=0,25 moli

2CH3-CH2-COOH + CaO -> (CH3-CH2-COO)2Ca + H2O

2 moli C3H6O2...................1 mol CaO
x moli...................................3 moli CaO

x=6 moli 

2CH3-CH2-COOH + CaCO3 -> (CH3-CH2-COO)2Ca + CO2 + H2O

2 moli C3H6O2...............1 mol CaCO3
x moli..............................0,25 moli CaCO3

x=0,5 moli 

n total = 6+0,5=6,5 moli acid

Cm=n/Vs => Vs=n/Cm=6,5/2= Calculeaza !!!