Question

How many grams of potassium chlorate of 80% purity is required to prepare 60 grams of oxygen?​

Answer 1

Answer:

In the question the mass has two significant figures; thus the mass of O2 released would be 4.0 g (two significant figures). 122.55 g of KClO3 yields 3*32 g of O2. 1 g of KClO3 would yield 0.783353733 g of O2. Thus 5 g of KClO3 would yield 5*0.783353733 g of O2 or 4.0 g O2.Feb 23, 2018

Answer 2

Answer:

here is your answer