STP তে 22.4L অক্সিজেন প্রস্তুত করতে কত গ্রাম পটাসিয়াম ক্লোরেট প্রয়োজন ?

The question asks for the amount of potassium chlorate (KClO₃) needed to produce 22.4L of oxygen (O₂) gas at standard temperature and pressure (STP). To understand why 81.73 grams is the correct answer, we can look at the chemical reaction involved:

Balanced Chemical Equation:

2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

This equation tells us that 2 moles of potassium chlorate produce 3 moles of oxygen. Since 1 mole of any gas at STP occupies 22.4 liters, we will need to calculate how many moles of KClO₃ are required to produce 1 mole of O₂ (which is 22.4L).

Calculating Moles of KClO₃:

    
• From the reaction, 2 moles of KClO₃ give 3 moles of O₂.
    
• To produce 1 mole of O₂, you need (2/3) moles of KClO₃.

Molar Mass of KClO₃:

    
• Potassium (K) = 39.1 g/mol
    
• Chlorine (Cl) = 35.5 g/mol
    
• Oxygen (O) = 16 g/mol (there are 3 O in KClO₃)
    
• Total Molar Mass = 39.1 + 35.5 + (3 × 16) = 122.55 g/mol

Calculating the Mass Required:

    
• To find the grams needed for (2/3) moles:
    
• Mass = Moles × Molar Mass
    
• Mass = (2/3) × 122.55 g/mol = 81.73 g

So, 81.73 grams of potassium chlorate is required to produce 22.4 liters of oxygen at STP, which confirms that the answer is correct.

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