KCl3+HCl-->Cl2+H2O+KCl
-stabiliti coeficientii
-calc. nr de moli de clor ce se pot obtine daca ar reactiona 2,5 g clorat de potasiu ,de puritate 98%.
-calc vol solutiei de acid clohidric de conc 36% si p=1,183 g/cm^3,care reactioneaza cu toata cantitatea de clorat de potasiu.

2

Răspunsuri

2014-06-06T10:18:19+03:00
2KCl _{3} +2HCl-->3Cl _{2} +H _{2} O+2KCl
1) p=p= \frac{mp}{m_{imp} } *100 
Nr. de moli=mp/masa molara
mp=mp=  \frac{m_{imp}*p}{100}
mp=(2,5*98)/100=2,45 g 
Nr. de moli= 2,45/74,5=0,03 moli KCl
Masa Molara= 39+35,5=74,5 g/mol KCl
x=(0,03*3)/2=0,045 moli Cl2

2) c%= md/ms *100 
ρ=ms/V
ms=(md*100)/c%
Masa Molara=1+35,5=36,5 g/mol HCl
y=(0,03*2)/2=0,03 moli HCl
Nr. de moli= md/masa molara => md = Nr.de moli * masa molara = 0,03*36,5 
md=1,095 g
ms= (1,095*100)/36=3,04 g
V= ms*ρ= 3,04*1,183= 3,59 ml HCl. 
Sper că nu am greșit la calcule, mai verifici tu o data.
Cel mai inteligent răspuns!
2014-06-06T13:21:52+03:00
KClO3 + 6 HCl ---> 3 Cl2 + 3 H2O + KCl
masa pura KClO3 = 2,5*98/100=2,45 g
122,5 g KClO3......6 moli HCl...3 moli Cl2
2,45 g KClO3.........x moli HCl.....y moli Cl2
x = 0,12 moli HCl
y = 0,06 moli Cl2 - rezultatul pt pct a
cM=ro*c%*10/masa molara HCl=36*1,1183*10/36,5=>cM=11,66795 M
cM=nr moli/volum=>volum HCl = 0,12/11,66
=>V HCl = 10,28 ml HCl - rezultat pt pct b 
4 3 4