Răspuns:
Explicație:
100g NH4Cl , p= 10,7 %
randament=75%
V amestec gaze
masa subst. solide ramase
- se afla masa pura de NH4 Cl
mp= mi . p : 100
mp= 100 . 10,7 : 100= 10,7 g NH4 Cl pura
-se afla masa de NH4 Cl care va reactiona [masa practica ]
mp= randam. . mt : 100
mp=75 . 10,7 : 100=8,025 g NH4Cl
-se afla g , molii , litrii de NH3
8,025g xg yg
NH4 Cl = NH3 + HCl
53,5g 17g 36,5g
x= 8,025 .17 : 53,5= 2,55 g NH3
n= 2,55g : 17g/moli=0,15 moli NH3
VNH3= 0,15 . 22,4l/moli=3,36 L
-se afla g, moli , L de HCl
y=8,025 . 36,5 : 53,5 =5,475 g HCl
n= 5,475g : 36,5g/moli=0,15 moli HCl
VHCl= 0,15moli. 22,4 L /moli=3,36 L HCl
-se afla V amestecului de gaze
V amestec= 3,36 + 3,36=6,72 L
===>masa subst . ramase [ m impurit . ]
100 - 10,7= 89,3 g imp.
----> masa subst. netransformata
10,7 g - 8,025 g= 1, 675 g
MNH4Cl= 14 + 4 + 35,5= 53,5 -------> 53,5g/moli
M NH3= 14 + 3= 17--------> 17g/moli
MHCl= 1 + 35,4=36,5--------> 36,5g/moli
