Question

Question #1bdae

Answer 1

`=31.0mL " of " O_2`

Explanation:

1. Write and balance the equation
   `2KClO->2KCl+O_2`
2. I assumed that the `mKClO=0.250g`. Per convention, convert it to moles. The molar mass of `KClO=(90.5g)/(mol)` which is obtainable from the periodic table; i.e.,
   `etaKClO=0.250cancel(gKClO)xx(1molKClO)/(90.5cancel(gKClO)`
   `etaKClO=0.002762mol`
3. Referring to the balanced equation, find the `etaO_2`, where:
   `=0.002762cancel(molKClO)xx(1molO_2)/(2cancel(molKClO))`
   `=0.001381molO_2`
4. At `STP`, mol of a gas will occupy a volume of 22.4L. This relationship will serve as the conversion factor to find the volume of `O_2` collected at this condition.
   `V_(O_2)=0.001381cancel(molO_2)xx(22.4L" of " O_2)/(1cancel(molO_2))`
   `V_(O_2)=0.03094L" of " O_2~~31.0mL" of "O_2`