Question

The reaction 2HgO(s) ↽⇀ 2Hg (l) + O2 (g), you produce 1 mole of Hg(I). What is your percentage yield?
A) 30%
B) 20%
C) 50%
D) 100%

Answer 1

The answer is going to be 50%

Answer 2

The percentage yield of Hg(l) is 50%. The correct option is C) 50%

From the question,

We are to determine the percentage yield.

Percentage yield is given by the formula,

$Percentage\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100\%$

From the given equation,

2 moles of HgO decompose to give 2 moles of Hg and 1 mole of O₂

That means,

Theoretical yield of Hg = 2 moles

Actual yield of Hg = 1 mole

Putting the parameters into the formula, we get

$Percentage\ yield\ of\ Hg= \frac{1}{2} \times 100\%$

$Percentage\ yield\ of\ Hg= 50\%$

Hence, the percentage yield of Hg(l) is 50%. The correct option is C) 50%

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