Question

m=153,125g K Cl 03
P=80%
Vo2=?

Answer 1

M(miu)=(39gK+35,5gCl+3x16gO)/mol=122,5g/mol

m,KClO3 pur rezulta din puritate
p/100= m,pur/m,impur
80/100= m,pur/153,125------> m,pur=122,5g
n(niu)= m/M
n= 122,5g/122,5g/mol= 1 molKClO3-

ecutia reactiei de descompunere este 
2mol.........................3mol
2KClO3------> 2KCl  + 3O2
1mol..........................1mol
V=nxV,m
V= 1molx22,4l/mol=...................... calculeaza !!!