Question

What mass of potassium chlorate must be decomposed to produce 2.4l of oxygen at 1.5 bar and 25°C​

Answer 1

Hi,

Answer:

Step 1 : We will write a balanced chemical equation

2 KClO₃ (s) → 2 KCl (s) + 3 O₂ (g)

Step 2: Calculating the moles of O₂

Pressure, P = 1.5 bar = 1.480 atm

Temperature, T = 25°C = 25 + 273 = 298 K

Volume, V = 2.4 L

Ideal Gas constant, R = 0.082057 L atm K⁻¹mol⁻¹

Using the Ideal Gas Law,

PV = nRT

Or, n = PV / RT = [1.480 * 2.4] / [0.082057 * 298] = 0.1452 moles of O₂

Step 3: Calculating the moles of KClO₃

From the above balanced equation we can see that  

2 moles of KClO₃ produces 3 moles of O₂

∴ No.of moles of KClO₃ that produces 0.1452 moles of O₂ = 2/3 * 0.1452 = 0.0968 moles

Step 4: Calculating the mass of KClO₃ / Potassium Chlorate

Molar mass of KClO₃ = 122.55 g

We have,  

No. of moles = [mass of KClO₃] / [Molar mass of KClO₃]  

Or, 0.0968 = mass of KClO₃ / 122.55

Or, Mass of KClO3 = 0.0968 * 122.55 = 11.86 g

Hence, 11.86 g of KClO₃ must be decomposed to produce 2.4 L of O₂ at the given condition.

Hope this helps!!!!!