| Lösungen |
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| 1) a) 1 mol H2SO4
= 2 x 1 + 32 + 4 x 16 g = 98 g
b) 0,5 mol NaOH = 0,5 x (23 + 16 + 1) g = 20 g c) 3 mol Al2O3 = 3 x ( 2 x 27 + 3 x 16) g = 306 g |
| 2) a) 21 g HNO3:
1 mol = 1 + 14 + 3 x 16 g = 63 g 21/63 = 0,33 mol
b) 28 g KOH : 1 mol = 39 + 16 + 1 g = 56 g 28/56 = 0,5 mol c) 2 g NaOH : 1 mol = 23 + 16 + 1 g = 40 g 2/40 = 0,05 mol |
| 3) a) 4 g H2
: 1 mol H2 = 2 g 4/2 = 2 mol 1
mol Gas: 22,4 l ==> 2 mol Gas: 44,8 Liter
b) 16 g O2 : 1 mol O2 = 32 g 16/32 = 0,5 mol 0,5 mol Gas: 11,2 Liter c) 48 g CH4 : 1 mol = 16 g 48/16 = 3 mol 3 mol Gas: 67,2 Liter |
| 4) a) 11,2 Liter H2-Gas:
22,4 Liter = 1 mol ==> 11,2 l = 0,5 mol 1 mol = 2 g ==>
0,5 mol = 1 g
b) 44,8 Liter O2 : 22,4 l = 1 mol ==> 44,8 l = 2 mol 1 mol O2 = 32 g ==> 2 mol = 64 g c) 33,6 Liter C3H8-Gas : 33,6 l = 1,5 mol 1 mol C3H8 = 44 g ==> 1,5 mol = 66 g |
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