Answer to Question #97366 in General Chemistry for Maria

Question

When heated, KClO3
decomposes into KCl
and O2.


2KClO3⟶2KCl+3O2


If this reaction produced 62.3 g KCl,
how many grams of O2
were produced?

Answer 1

Mass= Moles * Molecular weight

Moles of KCl -> 62.3/74.5 = 0.84 moles

Moles of KClO3=Moles of KCl (2KClO3⟶2KCl)

Moles of KClO3=0.84 moles

2KClO3⟶3O2

1 mole of KClO3 = 3/2 mole of O2

0.84 mole of KClO3 = 3/2 * 0.84 moles of O2 = 1.26 mole of O2

Mass of O2 = 1.26 * 16 = 20.16 g

20.16 g