Question

Oxygen is prepared by catalytic de-composition of potassium chlorate ($KClO_{3}$). Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen ($O_{2}$). If 2.4 mole of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed? (At mass K = 39, Cl = 35.5, O = 16)

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.

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

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

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

From the above reaction,

For 3 moles of ${ O }_{ 2 }$, we need $2\times 122.5g\quad of \quad KCl{ O }_{ 3 }$

For 2.4 moles of ${ O }_{ 2  }$, we need:

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

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