Question

raspundeti la 2.15, va rog

Answer 1

 a)
28 g X.... 0.5 mol atomi... 3.01115 * 10^23 atomi
x g X ........1 mol atomi......6.0223*10^23 atomi
x=56 g => Fe
Confirmam ca elementul este Fe
1 atom ... 26 e-
1 mol atomi ... 26 * 6.0223*10^23 e = 1.565798 * 10^25 e-
0.5 mol atomi ... 1.565798 * 10^25 e- /2 = 7.82899 *10^24 e- = 78.2899*10^23 e-
b) 1s2; 2s2 2p6; 3s2 3p6; 4s2; 3d6
c) grupa 8, perioada 4
d) n Cl2 =  21.3/71 =0.3 mol Cl 2
Fe + Cl2 => FeCl2
1 mol Fe... 1 mol Cl2
x mol Fe .... 0.3 mol Cl2 => x= 0.3 mol Fe

2Fe + 3Cl2 => 2FeCl3
2 mol Fe ... 3 mol Cl2
y mol Fe ...0.3 mol Cl2
y=0.2 mol Fe

d)
 26 e-... 1 mol 
2 e.........z => z=0.076 mol e-
1 mol Fe.... 0.076 mol e-
a mol Fe.....0.6 mol e- => a= 7.89 mol Fe => m Fe = 442.11 g Fe

26 e-.....1mol
3 e .........b => b= 0.115 mol e-
1 mol Fe....0.115 mol e-
c mol Fe.....0.6 mol e- => c= 5.21 mol Fe = 292.22 g Fe
f) Fe Valenta: II, III, Nr. Oxidare 2+, 3+