Question

if 2.4 mol of oxygen is needed how many grams of potassium chlorate must be used​

Answer 1

The catalytic decomposition reaction is carried out in the presence of catalyst called catalytic decomposition reaction.

Potassium dichlorate undergoes decomposition reaction to form potassium chloride and oxygen gas.

2KCl{ O }_{ 3 }(s)\rightarrow \quad 2KCl(s)+\quad 3{ O }_{ 2 }\quad (g)2KClO

3

(s)→2KCl(s)+3O

2

(g)

Molecular weight of KCl{ O }_{ 3 }KClO

3

39+\left( 35.5+3 \right) \times 16 = 122.5 g39+(35.5+3)×16=122.5g

From the above reaction,

For 3 moles of { O }_{ 2 }O

2

, we need 2\times 122.5g\quad of \quad KCl{ O }_{ 3 }2×122.5gofKClO

3

For 2.4 moles of { O }_{ 2 }O

2

, we need:

=\frac { 2\times 122.5 }{ 3 } \times 2.4\quad

3

2×122.5

×2.4

= 196 g of KCl{ O }_{ 3 }KClO

3

Hope it helps you dear ☺️