Question

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.

Answer 1

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