KClO3(s) produces KCl and O2 when heated. 2.34L of O2 was collected over water at 9878mmHg at 65°C. What mass of KClO3 reacted?

Question

How do I solve this AP Chemistry problem?

Answer 1

2KClO₃ ==∆==> 2KCl + 3O₂   ...   balanced equation

Assuming the pressure is 987.8 mm Hg (and not 9878 mm), then..

Find moles of O₂ produced. 

PV = nRT

P = 987.8 mm Hg - 187.5 mm Hg = 800.3 mm Hg (the 187.5 mm Hg that was subtracted is the vapor pressure of water at 65ºC)

V = 2.34 L

R = gas constant = 62.36 Latm/Kmole

T = 65 + 273 = 338K

n = PV/RT = (800.3 mm)(2.34 L)/(62.36 Lmm/Kmol)(338K)

n = 0.0889 moles O₂

Moles KClO₃ that reacted = 0.0889 moles O₂ x 2 moles KClO₃/3 moles O₂ = 0.0593 moles KClO₃

Mass KClO₃ that reacted = 0.0593 moles x 122.6 g/mole = 7.27 g (3 significant figures)