Question

7. You are given 2.263 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl and O2, 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g), and 632 mL of O2 is collected over water at 20 °C. The total pressure of the gases in the collection flask is 736 torr. What is the weight percentage of KClO3 in the sample? The formula weight of KClO3 is 122.55 g/mol. The vapor pressure of water at 20 °C is 17.5 torr.

Answer 1

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g),

V = 632ml    = 0.632L

T   = 20+273 = 293K

PT   = 736 torr

The vapor pressure of water at 20 °C (PH2O) = 17.5 torr.

PO2   = PT-PH2O

          = 736-17.5

        = 718.5torr

         = 718.5/760    = 0.945atm

PV = nRT

n    = PV/RT

       = 0.945*0.632/0.0821*293

        =0.025moles

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g),

3 moles of O2 produced from 2 moles of KClO3

0.025 moles of O2 produced from = 2*0.025/3   = 0.0167moles of KClO3

mass of KClO3   = no of moles * gram molar mass

                           = 0.0167*122.55   = 2.05g

percentage of KClO3 in the sample = 2.05*100/2.263   = 90.6%