What mass of potassium chlorate must be decomposed to produce `2.40 L` of oxygen at `0.82 "bar"` and `300 K`?

Explanation

The decomposition of potassium chlorate is represented as
` 2 KClO_(3)(s)to 2 KCl(s) + 3O_(2)(g)`
Step I Calculation of number of moles of oxygen
`P = 0.82` bar` ,` V = 2.40 L`, `T=300 K`
Using ideal gas equation, we get
`n=(PV)/(RT)=(0.82xx2.40)/(0.082xx300)=0.08 mol`
Step II: Calculation of amount of `KClO_(3)` from the chemical equation:
`3 mol` of oxygen is produced from `2 mol` of `KClO_(3)`.
So `0.08 mol` of oxygen is produced from `(2)/(3)xx0.08`
`=0.533 mol` of `KClO_(3)`
Now, `1 mol` of `KCl_(3)=122.5 g`
So, `0.0533 mol` of `KClO_(3)=122.5xx0.0533=6.533 g`