What is the mass of KClO₃ (s) required to liberate 22.4 dm³ oxygen at STP during thermal decomposition?
(Molar Mass of KClO₃ = 122.5 g/mol)
A. 122.5 g
B. 81.67 g
C. 10.25 g
D. 8.16 g
Answer (2)
To produce 22.4 dm³ of O₂ from KClO₃ decomposition at STP, we find that 81.67 g of KClO₃ is required. This is calculated based on the reaction stoichiometry and the molar mass of KClO₃. Therefore, the correct choice is B. 81.67 g.
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To solve this problem, we start by writing the balanced chemical equation for the thermal decomposition of potassium chlorate (KClO₃):
2 KClO 3 ( s ) → 2 KCl ( s ) + 3 O 2 ( g )
From the equation, we can see that decomposing 2 moles of KClO₃ produces 3 moles of O₂.
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 dm³. Therefore, to liberate 22.4 dm³ of oxygen gas, we need 1 mole of O₂.
From the balanced equation:
3 moles of O₂ are produced by 2 moles of KClO₃.
So, 1 mole of O₂ is produced by 3 2 moles of KClO₃.
Next, we calculate the mass of 3 2 mole of KClO₃:
The molar mass of KClO₃ is given as 122.5 g/mol.
Mass of 3 2 mole of KClO₃:
Mass = 3 2 moles × 122.5 g/mol Mass = 3 2 × 122.5 g Mass = 81.67 g
Therefore, the mass of KClO₃ required to liberate 22.4 dm³ of oxygen at STP is 81.67 g.
The correct answer is (B) 81.67 g.
