Question

In the reaction below, 22 g of H2S with excess

Answer 1

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%