Question

(2) Calculate the mass of KClO3 required to produce 6.72 L of O2 at
STP
2) Calculate the number of moles of O2 in the above volume and
also the number of molecules.

Answer 1

Answer:

24.5 gram required

Explanation:

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Answer 2

mass of KClO₃ would be 24.5 grams.

number of moles of oxygen gas = 0.3

number of molecules = 1.8069 × 10²⁴

Decomposition reaction of KClO₃ ....

2KClO₃ ⇔ 2KCl + 3O₂

here we see, 2 moles of KClO₃ decompose into 2 moles of KCl and 3 moles of oxygen gas (O₂).

given, volume of oxygen gas = 6.72 L

mole of oxygen gas at STP = 6.72/22.4 = 0.3

2 moles of KClO₃ = 3 moles of O₂

⇒2/3 × 0.3 moles of KClO₃ = 0.3 mol of O₂

⇒0.2 mole of KClO₃ = 0.3 mol of O₂

now mass of 0.3 mol of KClO₃ = 0.2 × molar mass of KClO₃

= 0.2 × 122.5 g/mol

= 24.5 grams.

now number of molecules = 0.3 × 6.023 × 10²³ = 1.8069 × 10²⁴

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