Opgave 1.
a] Na2CO3•10H2O
b] CaSO4•2H2O
Opgave 2.
a] FeCl3•6H2O(s) → Fe3+(aq) + 3 Cl-(aq) + 6 H2O(l)
b] Na2SO3•7H2O(s) → Na2SO3(s) + 7 H2O(g)
Opgave 3.
(massa 7 H2O)/(massa FeSO4•7H2O) x 100% = (7x18,105)/(151,91 + 7x18,105) x 100% = 45,36 massa%
Opgave 4.
Ba(OH)2•8H2O(s) → Ba2+(aq) + 2 OH-(aq) + 8 H2O(l)
0,250 L x 0,0010 mol/L = 2,5x10-4 mol OH-
2,5x10-4 / 2 = 1,25x10-4 mol Ba(OH)2•8H2O
De molaire massa van Ba(OH)2•8H2O is: 171,34 + 8x18,015 = 315,46 g/mol
1,25x10-4 mol x 315,46 g/mol = 0,039 gram
Antwoord: 39 mg bariumhydroxideoctahydraat.