Answer to Question #279583 in Chemistry for Mosab

Question

The decomposition of KClO

3

is commonly used to prepare small amounts of O

2

in the

laboratory:

2 KClO

3

s → 2 KCl s + 3 O

2

g

How many grams of O

2

can be prepared from 4.50 g of KClO

3

?

Answer 1

The decomposition of KClO₃ is commonly used to prepare small amounts of O₂ in the laboratory:

2 KClO_3(s) → 2 KCl_(s) + 3 O_2(g)

How many grams of O₂ can be prepared from 4.50 g of KClO₃?

According to the equation, n (O₃) = 3/2 x n (KClO₃).

n = m / M

M (O₂) = 32 g/mol

M (KClO₃) = 122.6 g/mol

n (KClO₃) = 4.50 / 122.6 = 0.04 mol

n (O₂) = 3/2 x n (KClO₃) = 3/2 x 0.04 = 0.06 mol

m (O₂) = n x M = 0.06 x 32 = 1.92 g