M.Ca(OH)2 = 40+(16+1)2 = 74 g/mol
M.HCl = 1+35,5 = 36,5 g/mol
M.CaCl2 = 111 g/mol
M.H2O = 1x2+16 = 18 g/mol
m g 10,5 g a g b g
Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
74 2x36,5 111 2x18
=> m= 74x10,5/2x36,5 = 10,64 g Ca(OH)2 necesara
=> a = 10,5x111/2x36,5 = 15,97 g CaCl2
=> b = 10,5x2x18/2x36,5 = 5,178 g H2O
intrat = 10,64 + 10,5 = 21,14 g
iesit = a+b = 15,97+5,78 = 21,14 g
deci , legea conservarii masei se respecta