Question

6,09 g MgCl(doi)·xH(2)O.La sol obtinuta s-a adaugat exces de AgNo(3) si sa format 8,61 g de precipitat. Determinati formula cristalo-hidratului

Answer 1

MgCl2 + 2AgNO3 => Mg(NO3)2 + 2AgCl

95g MgCl2....................2×143,5g AgCl
xg MgCl2........................8,61g AgCl

x = 2,85g MgCl2

m H2O = 6,09-2,85 = 3,24g

95+18xg.................18xg H2O
6,09g...................3,24g

307,8= 51,3x

x = 6

F.M => MgCl2×6H2O