Question

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 %.

Answer 1

24.1 %

Step-by-step explanation:

This is the reaction of oxygen between hydrogen sulfide.

The equation is:

2H₂S + O₂ → 2S + 2H₂O

As the oxygen is the excess, limting reagent is the H₂S.

We convert the mass to moleS: 22g / 34.06 g/mol =

0.646 moles

Ratio is 2:2. 2 moles of sulfide can produce 2 moles of sulfur.

Then, 0.646 moles of sulfide will produce 0.646 moles of S.

We convert the moles to mass: 0.646 mol . 32.06 g/mol =

20.71 g

That's the theoretical yield.

% yield = (produced yield / theoretical yield) . 100

% yield = (5 g/ 20.71g) . 100 = 24.1 %