masa pura = 49 g
puritate p% = 80%
daca 49 g ......................... 100% impur
cat a g ............................... 80% pur
a = 39,20 g KClO3 pur
39,20 g m g
2KClO3 --> 2KCl + 3O2
2x122,5 2x74,5
=> m = 2x74,5x39,2/2x1225,
= 23,84 g s-ar obtine la randament la 100%, dar la 75% cat ar fi..
23,84 g ...................... randament 100%
b g ................................... 75%
=> b = 17,88 g KCl
