Question

400 g solutie HCl cu c=20% neutralizeaza o solutie de NaOH cu c=15%
1. masa de sare rezultata
2.masa solutiei de NaOH consumata
3.masa apeoi din solutia finala
4.concentratia solutiei finale.

Answer 1

36.5g/mol     40 g/mol     55.84 g/mol     18g/mol
      HCl     +    NaOH =>      NaCl       +    H2O
      80 g            a g                b g                d  g 
ms HCl= 400 g
c HCl=20 %
md HCl= c x ms/ 100 => md HCl = 20 x 400/100 = 80 g HCl pur

c NaOH= 15%
a= 80 x 40/36.5 = 87.67 g NaOH pur (md)
2) ms NaOH = md x100/c = 87.67 x 100/ 15 =584.46 g sol NaOH 15%

1)  b=80 * 55.84 /36.5 = 122.4 g NaCl
3) ms = md + m H2O=> m H2O = ms - md
m H2O ( din sol HCl) = 400 g sol. - 80 g HCl = 320 g H2O
m H2O (din sol NaOH) = 584.46 g sol. - 87.67 g  NaOH = 496.79 g H2O
m H2O ( din reactie adica "d") = 80 x 18 / 36.5 = 39.45 g H2O
m H2O total = 320 g + 496.79 g + 39.45 g= 856.24 g H2O

4) ms final= m NaCl + m H2O final = 122.4 g NaCl + 856.24 g H2O= 978.64 g sol.
md = 122.4 g NaCl
c= md/ms x 100 => c= 122.4 / 978.64  x 100 = 12.51 %