Question #198971

2KCLO_{3}→ 2KCl + 3O_{2(g) }

1.5 mole of KCLO_{3} was decompose during an experiment in laboratory 40 dm^{3}

of O_{2} was produced find out

percentage yield of O_{2. }

Expert's answer

**Solution:**

Balanced chemical equation:

2KClO_{3}(aq) → 2KCl(aq) + 3O_{2}(aq)

According to the equation above: n(KClO_{3})/2 = n(O_{2})/3

Therefore,

n(O_{2}) = 3 × n(KClO_{3}) / 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 O_{2} occupies:

(2.25 mol) × (22.4 L / 1 mol) = 50.4 L O_{2}

50.4 L O_{2} - theoretical yield

40 dm^{3} O_{2 }= 40 L O_{2 }- practical yield

Therefore,

%O_{2 }= (40 L / 50.4 L) × 100% = 79.365% = 79.37%

**%O**_{2 }**= 79.37%**

**Answer: The percentage yield of O**_{2}** is 79.37%**

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