A sample of KClO₃ on decomposition yielded 448 mL of oxygen gas at NTP.

• Calculate (i) weight of oxygen produced
• (ii) weight of KClO₃ originally taken
• (iii) weight of KCl produced.

(K = 39, Cl = 35.5, and O = 16)

To solve the problem, we need to analyze the decomposition reaction of potassium chlorate (KClO₃), which can be represented as:

Decomposition Reaction

The balanced equation for the decomposition of KClO₃ is:

2 KClO₃ → 2 KCl + 3 O₂

Step 1: Calculate the weight of oxygen produced

At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters (or 22400 mL). Given that 448 mL of oxygen gas is produced, we can find the number of moles of O₂:

• Number of moles of O₂ = Volume of O₂ / Molar volume = 448 mL / 22400 mL/mol = 0.02 moles

Next, we calculate the weight of oxygen produced:

• Molar mass of O₂ = 2 × 16 g/mol = 32 g/mol
• Weight of O₂ = Number of moles × Molar mass = 0.02 moles × 32 g/mol = 0.64 g

Step 2: Determine the weight of KClO₃ originally taken

From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂. Therefore, we can find the moles of KClO₃ that produced 0.02 moles of O₂:

• Moles of KClO₃ = (2/3) × Moles of O₂ = (2/3) × 0.02 = 0.01333 moles

Now, calculate the weight of KClO₃:

• Molar mass of KClO₃ = 39 + 35.5 + (3 × 16) = 122.5 g/mol
• Weight of KClO₃ = Moles × Molar mass = 0.01333 moles × 122.5 g/mol = 1.63 g

Step 3: Calculate the weight of KCl produced

From the balanced equation, 2 moles of KClO₃ yield 2 moles of KCl. Thus, the moles of KCl produced will be equal to the moles of KClO₃ used:

• Moles of KCl = 0.01333 moles

Now, calculate the weight of KCl:

• Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
• Weight of KCl = Moles × Molar mass = 0.01333 moles × 74.5 g/mol = 0.99 g

Summary of Results

• Weight of oxygen produced: 0.64 g
• Weight of KClO₃ originally taken: 1.63 g
• Weight of KCl produced: 0.99 g