Question

considerand ca s-au folosit 200 g de clorura calciu 11,1 imputritati si ca masa subst b obtinute se imparte in mod egal in trans urmatoare.

Answer 1

CaCl2 + 2 H2O = Ca(OH)2 + 2 HCl
m CaCl2 impur = 200 g 
p= 100 -11.1 = 88.9 %
m CaCl2 pur = 200 * 88.9/100 = 177.8 g CaCl2 pur
a)
111g/mol CaCl2 ...2 x 36.5 g/mol HCl
177.8 g CaCl2 .......a g HCl
a = 177.8 x 2 x 36.5 / 111 = 116.93 g  HCl
n HCl = 116.93 g / 36.5 g/mol = 3.206 mol HCl
V HCl = 3.206 mol x 22.4 L/mol = 71.76 L HCl (cn)

b)
111g/mol CaCl2 .... 74.1 g/mol Ca(OH)2
177.8 g CaCl2 .........b g Ca(OH)2
b = 177.8 x 74.1 / 111 =118.7 g Ca(OH)2

c)
c H2SO4 = 24.5 %
m Ca(OH)2 = 118.7 g
d H2SO4 = 1.841 g/ml
Ca(OH)2 + H2SO4 => CaSO4 + 2 H2O

74.1 g/mol Ca(OH)2 ... 98 g/mol H2SO4 
118.7 g Ca(OH)2..............c g H2SO4
c = 118.7 x 98 / 74.1 = 156.98 g H2SO4 (md)
ms H2SO4 = md x 100 / c
ms H2SO4 = 156.98 x 100 / 24.5 = 640.73 g sol H2SO4 24.5 %
d= ms/Vs
Vs=ms/d 
Vs H2SO4 = 640.73/1.841 = 348.1 ml sol H2SO4 24.5%

d)m Ca(OH)2 = 118.7 g Ca(OH)2 (vezi b)
n Ca(OH)2 = 118.7 g/ 64.1 g/mol = 1.85 mol Ca(OH)2

e) gips = CaSO4
74.1 g/mol Ca(OH)2 ... 136.14 g/mol CaSO4
118.7 g Ca(OH)2  ...........d g CaSO4
d = 118.7 x 136.14 / 74.1 = 218.1 g CaSO4 anhidru