Thermal decomposition of x g of `KClO_(3)` produces 760 mL `O_(2)`, which is collected over water at `27^(@)C` and 714 mm Hg pressure. Find the value of x. [Given tht at `27^(@)C`, aqueous tension=26 mm Hg and atomic weights of K=39, Cl=35.5, O=16]

Explanation

Actual pressure `O_(2)` collected over water
=atmospheric pressure-aqueous tension
`=(714-26)=688mm" "Hg=(688)/(760)=0.9052` atm
According to the given conditions volume of `O_(2)`
`=760mL=0.76L and T=(273+27)K=300K`
If the number of moles of `O_(2)` gas be n then,
`n=(PV)/(RT)=(0.9052atmxx0.76L)/(0.0821L*atm*mol^(-1)*K^(-1)xx300K)=0.028` mol
Equation for the thermal decomposition of `KClO_(3)` is,
`2KClO_(3)(s) to 2KCl (s)+3O_(2)(g)`
Therefore, 3mol `O_(2)(g)-=2" mol "KClO_(3)=2xx122.5g`
=245 g `KClO_(3)`
`therefore 0.028mol" "O_(2)(g)-=(245)/(3)xx0.028g=2.286g" "KClO_(3)`
`therefore x=2.286g`