Question

How many grams of potassium chlorate decompose to potassium chloride and 363 mL of O2 at 128°C and 740. torr? 2 KClO3(s) → 2 KCl(s) + 3 O2 (g)

Answer 1

We know from gas law equation

P * V = n * R * T --------(1)

P = Pressure in atm, V = Volume in Litre

n = number of moles , R = gas constant =0.082 atm L mol-1K-1

T = Temperure in Kelvin

V = 363 ml = 363 L / 1000 = 0.363 L

T = 128 oC = (128 + 273)K = 401 K

P = 740 torr = 0.00132 * 740 atm = 0.977 atm

From (1)

0.977 atm * 0.363 L = n * 0.082 atm L mol-1K-1 * 401K

0.3546 atm L = n * 32.882 atm L mol-1

n = 0.3546 / 32.882 mol^-1 = 0.0108 mol

From Reaction, 3.0 mol of O2 decompose from 2.0 mol of KClO3 so 0.0108 mol of O2 will decompose from 2.0 * 0.0108 mol / 3.0 = 0.0072 mol of KClO3

Molar mass of KClO3 = 122.55 g/mol

Mass of KClO3 decomposed = 122.55 g/mol * 0.0072 mol = 0.88 g