Answer to Question #198971 in Chemistry for Hsn

Question #198971

2KCLO3→ 2KCl + 3O2(g)


1.5 mole of KCLO3 was decompose during an experiment in laboratory 40 dm3

of O2 was produced find out 

percentage yield of O2.

1
Expert's answer
2021-05-31T02:08:37-0400

Solution:

Balanced chemical equation:

2KClO3(aq) → 2KCl(aq) + 3O2(aq)

According to the equation above: n(KClO3)/2 = n(O2)/3

Therefore,

n(O2) = 3 × n(KClO3) / 2 = (3 × 1.5 mol) / 2 = 2.25 mol


At Standard Temperature and Pressure (STP), one mole of any gas occupies a volume of 22.4 L.

Thus 2.25 mol of O2 occupies:

(2.25 mol) × (22.4 L / 1 mol) = 50.4 L O2


50.4 L O2 - theoretical yield

40 dm3 O2 = 40 L O2 - practical yield

Therefore,

%O2 = (40 L / 50.4 L) × 100% = 79.365% = 79.37%

%O2 = 79.37%


Answer: The percentage yield of O2 is 79.37%

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