Question

a)Numarul de molecule de apa continute in 1,8 g apa
b)Numarul de molecule de H2 continute in 67,2 L H2
c)Masa de Al O3 continuta in 5 molecule de oxid de Al
d)Numarul de moli de acid clorhidric care contine 18,066x10 la puterea 22 molecule
e)Numarul de molecule continut in 2,24 L Cl2

Answer 1

a) n=m/M=1.8/18=0.1 moli H2O

1mol H2O..............6.022*10^23 molecule
0.1 moli H2O........x=6.022*10^22 molecule


b) n=V/Vm=67.2/22.4=3 moli H2

1 mol H2........6.022*10^23 molecule
3moli H2............x=18.066*10^23 molecule

c) 1mol Al2O3.......6.022*10^23 molecule
x=0.83 moli..........................5 molecule

mAl2O3=0.83*102=84.66g Al2O3


d) 1 mol HCl...........6.022*10^23 molecule
x=0.3 moli...................1.8066*10^23 molecule


e) n=V/Vm=2.24/22.4=0.1 moli Cl2

1mol Cl2..........6.022*10^23 molecule
0.1 moli Cl2.....x=6.022*10^22 molecule