1. M(NaCl) = 58,443 g/mol; n = m / M = 0,1711 mol = 171 mmol
2. M(Fe) = 55,847 g/mol; m = n * M = 0,005 * M = 0,2792 g = 279 mg
3. 2 Cl in 1 CaCl₂; n(Cl) = 2 * 0,8 = 1,6 mol
4. 2 Fe in 1 Fe₂O₃; 50 mmol Fe in 25 mmol Fe₂O₃
5. M(S) = 32,06 g/mol; M(Na₂S) = 78,04 g/mol; 1 S in 1 Na₂S;
   3 g Na₂S: 0,03844 mol Na₂S = 0,03844 mol S = 1,234 = 1,23 g S
6. M(Al) = 26,9815 g/mol; m(Al₂(SO₄)₃) = 342,14 g/mol; 2 Al in 1 Al₂(SO₄)₃.
   0,5 g = 0,01853 mol Al in 0,009266 mol Al₂(SO₄)₃ = 3,17 g.
   Zum Vergleich mit tabellierten "Umrechnungsfaktoren":
   Al₂(SO₄)₃ aus Al₂O₃: 3,3556; Al₂O₃ aus Al: 1,8895; gesamt (Produkt): 6,340;
   0,5 g * 6,340 = 3,17 g.
7. M(KClO₄) = 138,549 g/mol; 4 O in 1 KClO₄; 2 g KClO₄ = 0,01443 mol: 0,05774 mol = 57,7 mmol O
8. M(BaCl₂) = 208,24 g/mol; M(KCl) = 74,551 g/mol; 2 Cl in 1 BaCl₂; 2 g BaCl₂ = 0,009604 mol: 0,01921 mol Cl;
   1 KCl enthält 1 Cl, also gleiche Stoffmenge KCl; m(KCl) = 0,01921 * 74,551 = 1,432 = 1,43 g
9. 2 mol Al₂O₃ enthalten 6 mol O; 1 O in 1 BaO; also n(BaO) = 6 mol
10. M(CaBr₂) = 199,89 g/mol; M(Br) = 79,904 g/mol; 3 g CaBr2 = 0,0150 mol; doppelte Stoffmenge AlBr₃ = 0,0300 mol.
    2 Br in 1 CaBr₂ und 3 Br in 1 AlBr₃: 0,0150 * 2 + 0,0300 *3 = 0,120 mol Br = 9,59 g.

ZURÜCK: AUFGABEN 1