When heated, potassium chlorate, KClO3, melts and decomposes to potassium chloride and diatomic oxygen.?

[匿名]

2017-03-01 12:49 pm

a) What is the theoretical yield of O2 from 6.28g KClO3?
b) If 1.31g of O2 is obtained, what is the percent yield?

I balanced it to:
2KClO3 ---> 2KCl + 3O2

I tried finding the mass of O2 by converting the given mass of KClO3 to moles of O2, then to grams of O2.

I keep getting the theoretical yield wrong, therefore I can't solve part b. Someone help please!

回答 (1)

不用客氣

2017-03-01 1:03 pm

✔ 最佳答案

a)
Molar mass of KClO₃ = (39.0 + 35.5 + 16.0×3) g/mol = 122.5 g/mol
No. of moles of KClO₃ = (6.28 g) / (122.5 g/mol) = 0.05127 mol

2KClO₃ → 2KCl + 3O₂
Mole ratio KClO₃ : O₂ = 2 : 3
Maximum no. of moles of O₂ produced = (0.05127 mol) × (3/2) = 0.07691 mol

Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
Theoretical yield of O₂ = (0.07691 mol) × (32.0 g/mol) = 2.46 g

Alternative method :

Mass of 1 mole of KClO₃ = (39.0 + 35.5 + 16.0×3) g = 122.5 g
Mass of O in 1 mole of KClO₃ = 16.0 × 3 g = 48.0 g

All O in KClO₃ are converted to O₂.
Theoretical yield of O₂ = (6.28 g) × (48.0/122.5) = 2.46 g

b)
Percent yield = (1.31/2.46) × 100(%) = 53.3(%)

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