Question

How many grams of KClO3 are required to produce 128grams of O2

Answer 1

2KClO3 ==> 2 KCl + 3O2

393g O2 x 1 mole/32 g = 12.28 moles O2

moles KClO3 needed = 12.28 mol O2 x 2 mol KClO3/3mol O2 = 8.19 mol

grams KClO3 = 8.19 moles x 123g/mol = 1007 g

Answer 2

hello friends

my answer 326.4 gram something

approximately =327 gram right answer