1)Calculati concentratia acidului clorhidric 300 g solutie stiind ca acesta neutralizeaza 120 g NaOH.
NaOH + HCl ---> NaCl + H2O
40 g NaOH.........36,5 g HCl
120 g NaOH.........x g HCl x = 109,5 g HCl=md
c%=md/ms * 100 = 109,5/300 * 100 = 36,5%
2)Calculati masa de NaOH stiind ca nr de moli de HCl este egal cu 4.
NaOH + HCl ---> NaCl + H2O
1 mol NaOH...........1 mol HCl
x moli NaOH...........4 moli HCl x = 4 moli NaOH
nr moli = masa/masa molara =>masa = 4 * 40 = 160 g
3)Sa se calculeze masa solutiei de hidroxid de sodiu de concentratie 40% stiind ca HCl reactioneaza cu CaCO3 formand 222 g CaCl2
2 HCl + CaCO3 ---> CaCl2 + H2O + CO2
2 * 36,5 g HCl.........111 g CaCl2
x g HCl...................222 g CaCl2 x = 4*36,5 = 146 g HCl
HCl + NaOH ---> NaCl + H2O
36,5 g HCl.........40 g NaOH
146 g HCl...............y g NaOH y = 160 g NaOH = md
c%=md/ms * 100 => ms = 160/40 * 100 = 400 g solutie NaOH
4)Din reactia NaOH cu CuSO4 s-au obtinut 80 g CuSO4,iar concentratia solutiei NaOH = 20%.HCl 36,5 % reactioneaza cu AgNO3 si formeaza 287 g precipitat.Calculati raportul masic ms HCl : ms NaOH
2 NaOH + CuSO4 ---> CuCl2 + Na2SO4
2 * 40 g NaOH..........160 g CuSO4
x g NaOH.................80 g CuSO4 x = 40 g NaOH = md
c%=md/ms * 100=>ms = 40/20 * 100 = 200 g solutie NaOH
HCl + AgNO3 ---> AgCl (precipitat)+ HNO3
36,5 g HCl.........143,5 g AgCl
y g HCl.............287 g AgCl y = 73 g HCl = md
c%=md/ms * 100 =>ms = 73/36,5 * 100 = 200 g solutie HCl
raportul masic = ms HCl : ms NaOH = 200 : 200 => raportul este de 1 la 1
