Question

(3) Potassium chlorate (KCIO3) is known to decompose into potassium chloride (KCl) and oxygen gas (O2) upon heating: 2KCIO3 → 2KCl + 302 Suppose 10.0g of potassium chlorate is heated to complete decomposition. What is the volume of oxygen gas produced at room temperature (25°C) and atmospheric pressure (1.00 atm)? {R = 0.0821 L'atm/K mol}

Answer 1

Moles of KClO3 = mass/molar mass = 10/122.55 = 0.0816 moles

From balanced reaction;

2 mole KClO3 produces 3 moles of O2

So, 0.0816 mole KClO3 will produce = 0.0816 * 3 / 2 = 0.1224 moles of O2

Moles of O2 (n) = 0.1224

Temperature (T) = 25°C = 25+273 = 298 K

Gas constant (R) = 0.0821 L atm/mol K

Pressure (P) = 1 atm

Let volume is V liters

Using ideal gas equation;

PV = nRT

V = nRT/P = 0.1224 * 0.0821 * 298 / 1 = 2.995 liters OR 3 liters (approx) ....Answer

Let me know if any doubts.