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In the reaction below, 22 g of H2S with excess
O2 produced 5 g of sulfur.
? H2S + ? O2 → ? S + ? H2O .
What is the percent yield of sulfur?
Answer in units of %.

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pstnonsonjoku

Answer:

24%

Explanation:

From the question, the limiting reactant is H2S.

The reaction equation is;

2H2S + O2 → 2S + 2H2O

Number of moles of H2S reacted = 22g/34 g/mol = 0.647 moles

According to the reaction equation;

2 moles of H2S yields 2 moles of sulphur

0.647 moles of H2S yields 0.647 moles of sulphur

So;

Theoretical yield of sulphur = 0.647 moles * 32 g/mol = 20.7 g

Actual yield = 5 g

% yield = actual yield/theoretical yield * 100

% yield =5 g/20.7 g * 100

% yield = 24%

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