Răspuns :
Răspuns:
Explicație:
365g sol. HCl , c=10%
react. cu AgNO3
masa de precipitat=?
-se afla md sol. de HCl
md= c . ms / 100
md= 10 . 365 / 100= 36,5 g HCl
36,5g xg
HCl + AgNO3= AgCl↓ + HNO3
36,5g 143,5g
x= 36,5 . 143,5 / 36,5= 143, 5g AgCl
MHCl=1 +35,5=36,5-----36,5g/moli
MAgCl= 108 + 35,5= 143,5------ 143,5g/moli
ms=365g
C=10%
md=?
md=Cxms/100=10×365/100=36,5g
MHCl=1+35,5=36,5g/mol
MAgCl=108+ 35,5=143,5g/mol
36,5g....143,5g
AgNO3 + HCl => AgCl + HNO3
36,5g.....xg
x=36,5×143,5/36,5=143,5g AgCl.