Question

 

Answer 1

nKCl03 = 24,5/ 122,5 = 0,2 mol

2KCl03 -->2KCl + 3O2

0,2--------------------->0,3

nFe= 30/56 = 0,54 mol

2O2 + 3Fe -----> Fe304

0,3------------------->0,15

mFe304 = n.M = 0,15 . 232 = 34,8 gam

Answer 2

2KClO₃-to>2KCl+3O₂

0,2----------------------0,3

3Fe+2O₂-to>Fe₃O₄

0,3---------------0,15

n _KClO3=\(\dfrac{24,5}{122,5}\)=0,2 mol

nFe=\(\dfrac{30}{56}\)=0,53 mol

=>Fe dư 

=>m _Fe3O4=0,15.232=34,8g