Question

4. What volume of gas at STP would be generated by reaction of a mass of 0.25 g of KCIO3? Look in your lab manual prelab for the reaction. I

Answer 1

Solution:-

2KClO₃ $\rightarrow$ 2KCl + 3O₂

It is clear from balanced chemical equation that 2 moles of KClO₃ produces 3 moles of oxygen gas. Or, 1 mole of KClO₃ produce (3/2) moles of oxygen gas.

Number of moles in 0.25 g of KClO₃ = Mass ÷ molar mass

Molar mass of KClO₃ = (1 × molar atomic mass of K + 1 × molar atomic mass of Cl + 3 × molar atomic mass of O) = (1 × 39.098 + 1 × 35.454 + 3 × 15.999) = 122.55 g/mole

Number of moles of KClO₃ = (0.25 g) ÷ (122.55 g/mol) = 2.04 × 10^-3 mole

2.04 × 10^-3 moles of KClO₃ produces = (3/2) × (2.04 × 10^-3 moles) = 3.06 × 10^-3 moles of O₂

Now,

1 mole of gas at STP = 22.4 L

Thus, 3.06 × 10^-3 moles of O₂ = (3.06 × 10^-3) × 22.4 = 0.06854 L = 68.54 mL

Hence, the volume of oxygen gas generated = 0.06854 L or 68.54 mL

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Answer 2

Solution:-

2KClO₃ $\rightarrow$ 2KCl + 3O₂

It is clear from balanced chemical equation that 2 moles of KClO₃ produces 3 moles of oxygen gas. Or, 1 mole of KClO₃ produce (3/2) moles of oxygen gas.

Number of moles in 0.25 g of KClO₃ = Mass ÷ molar mass

Molar mass of KClO₃ = (1 × molar atomic mass of K + 1 × molar atomic mass of Cl + 3 × molar atomic mass of O) = (1 × 39.098 + 1 × 35.454 + 3 × 15.999) = 122.55 g/mole

Number of moles of KClO₃ = (0.25 g) ÷ (122.55 g/mol) = 2.04 × 10^-3 mole

2.04 × 10^-3 moles of KClO₃ produces = (3/2) × (2.04 × 10^-3 moles) = 3.06 × 10^-3 moles of O₂

Now,

1 mole of gas at STP = 22.4 L

Thus, 3.06 × 10^-3 moles of O₂ = (3.06 × 10^-3) × 22.4 = 0.06854 L = 68.54 mL

Hence, the volume of oxygen gas generated = 0.06854 L or 68.54 mL

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