1.5 g KClO3 decomposes in two ways :--

2KClO3---> 2KCl+3O2

4KClO3-----> 2KClO4+KCl

If the gm wt of KClO4 is 0.395g, Find the Volume of O2 at STP

To solve the problem of finding the volume of oxygen gas (O2) produced from the decomposition of potassium chlorate (KClO3), we first need to analyze the reactions provided and determine how much O2 is generated from the given mass of KClO3. Let's break this down step by step.

Understanding the Reactions

We have two decomposition reactions for KClO3:

• 2 KClO3 → 2 KCl + 3 O2
• 4 KClO3 → 2 KClO4 + KCl

However, for our calculations, we will focus on the first reaction, as it directly produces O2 gas.

Calculating Moles of KClO3

The first step is to determine how many moles of KClO3 are present in 1.5 grams. The molar mass of KClO3 can be calculated as follows:

• Potassium (K): 39.1 g/mol
• Chlorine (Cl): 35.5 g/mol
• Oxygen (O): 16.0 g/mol (3 atoms)

So, the molar mass of KClO3 is:

Molar mass of KClO3 = 39.1 + 35.5 + (3 × 16.0) = 122.6 g/mol

Now, we can calculate the number of moles of KClO3 in 1.5 g:

Moles of KClO3 = mass (g) / molar mass (g/mol) = 1.5 g / 122.6 g/mol ≈ 0.01224 moles

Using Stoichiometry to Find Moles of O2

From the balanced equation, we see that 2 moles of KClO3 produce 3 moles of O2. Therefore, we can set up a proportion to find the moles of O2 produced from our calculated moles of KClO3:

Moles of O2 = (3 moles O2 / 2 moles KClO3) × moles of KClO3

Moles of O2 = (3/2) × 0.01224 ≈ 0.01836 moles

Calculating Volume of O2 at STP

At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 liters. To find the volume of O2 produced, we multiply the moles of O2 by the volume per mole at STP:

Volume of O2 = moles of O2 × 22.4 L/mol

Volume of O2 = 0.01836 moles × 22.4 L/mol ≈ 0.4118 liters

Final Result

Thus, the volume of oxygen gas produced from the decomposition of 1.5 g of KClO3 at STP is approximately 0.4118 liters, or about 411.8 mL.