Dioxygen gas can be prepared by heating potassium chlorate `(KClO_(3))` with magnese dioxide `(MnO_(2))` as a catalyst:
`2KClO_(3)(S) overset(Delta) rarr 2KCl(s) + 3O_(2)(g)`
The gas is colleted over water. If `156 mL` of gas is collted at `20^(@)C` and `769 mm Hg`, what is the mass of dioxygen collected ?
Aqueous tension at `20^(@)C` is `17 mmHg`.
Strategy : The gaseous sample collected over water is dioxygen mixed with water vapour. To get mass of `O_(2)`, first find its partial pressure in the mixture using Dalton's law (step 1). Then calculate the moles of `O_(2)` from the ideal gas law (step 2). Finally, obtain the mass of `O_(2)` from its moles (step 3).

Explanation

According to Dalton's law,
`p_(t o tal) = p_(O_(2))+ p_(H_(2)O)`
`:. p_(O_(2)) = p_(t o tal) - p_(H_(2)O)`
`= (769 mm Hg) - (17 mm Hg)`
`= 752 mm Hg`
Step 2 : `p = 752 mm Hg xx (1 atm)/(760 mm Hg) = 0.989 atm`
`V = 156 mL xx (1L)/(1000 mL) = 0.156 L`
`T = (20 + 273)K = 293 K`
According to the ideal gas law,`pV = nRT`, we have
`n = (pV)/(RT) = ((0.989 atm)(0.156 L))/((0.821 (L atm)/(K mol))(293 K)) = 0.00641 mol`
Step 3: Converting moles of `O_(2)` to grams of `O_(2)`
`n_(O_(2)) = (Mass (O_(2)))/(Molar mass(O_(2)))`
`:. Mass(O_(2)) = (n_(O_(2))) (Molar mass O_(2))`
`= (0.00641 mol) (32 g mol^(-1))`
`= 0.21 g`