Question

You are given 2.674 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 758 mL of O2 is collected over water at 21 °C. The total pressure of the gases in the collection flask is 725 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 21 °C is 18.7 torr

Answer 1

Balanced equation:
2 KClO₃(s) ===> 2 KCl(s) + 3 O₂(g)
Reaction type: decomposition

Let us calculate the moles of O2

PV= nRT

P = Pressure in atm            V= Volume in Liter

n = no of moles             R = 0.0821 L atm K-1 Mol-1

T = Temperature in Kelvin

P = 725 Torr - 18.7 Torr = 706.3 / 760 = 0.929 atm

V = 758 ml = 0.758 Liter

T = 273 +21 = 294 K

n = 0.929 x 0.758 / 0.0821 x 294 = 0.02918 Moles

Moles of KClO3 reacted = 0.019453 Moles

Mass of KClO3 reacted =  0.019453 x 122.55 =  2.383 gm

weight percentage of KClO3 = 2.383 x 100 / 2.674 = 89.11 %