Question

Potassium chlorate (KClO3 molar mass 122.5 g/mol) decomposes into oxygen gas and potassium chloride. When 3.25 grams of potassium chlorate decomposes at 610 torr and 34*C, what is the volume of oxygen gas produced? Remember to convert torr to atm and Celcius to Kelvin. 2 KClO3 --> 3 O2(g) + 2 KCl (s).

Answer #1

2 KClO_{3} (s) -----------> 2 KCl (s) + 3
O_{2} (g)

Mass of KClO3 given = 3.25 g.

Molar mass of KClO_{3} = 122.5 g/mol

Number of moles of KClO_{3} = mass / molar mass = 3.25 /
122.5

n = 0.0265 mol

From the balanced equation,

2 mol of KClO3 gives 3 mol O2

then, 0.0265 mol of KClO3 gives 3 * 0.0265 / 2 = 0.0398 mol

Pressure , P = 610 torr = 610 / 760 atm = 0.803 atm

temperature , T= 34 degree C = 34 + 273.15 K = 307.15 K

gas constant, R = 0.0821 L.atm.K-1.mol-1

Ideal gas equation,

P V = n R T

V = 0.0398 * 0.0821 * 307.15 / 0.803

V = 1.25 L

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