In the reaction below, 23 g of H2S with excess

O2 produced 9 g of sulfur.
? H2S + ? O2 → ? S + ? H2O .
What is the percent yield of sulfur?
Answer in units of %

To calculate the percent yield of sulfur, we need to compare the actual yield (9 g of sulfur) with the theoretical yield, which is the amount of sulfur that would be produced if the reaction went to completion using the given quantities of reactants.

To find the theoretical yield, we first need to determine the balanced chemical equation for the reaction:

H2S + O2 → S + H2O

Looking at the equation, we can see that the stoichiometric ratio between H2S and S is 1:1, which means that for every 1 mole of H2S that reacts, 1 mole of S is produced.

Given that the molar mass of H2S is approximately 34 g/mol, we can convert the mass of H2S (23 g) to moles by dividing it by the molar mass:

23 g H2S × (1 mol H2S / 34 g H2S) = 0.676 mol H2S

Since the stoichiometric ratio between H2S and S is 1:1, the theoretical yield of sulfur would be 0.676 mol.

To convert the theoretical yield from moles to grams, we multiply by the molar mass of sulfur (approximately 32 g/mol):

0.676 mol S × (32 g S / 1 mol S) = 21.632 g S

So, the theoretical yield of sulfur is 21.632 g.

Finally, to calculate the percent yield of sulfur, we can use the following formula:

Percent Yield = (Actual Yield / Theoretical Yield) × 100%

From the given information, the actual yield of sulfur is 9 g.

Plugging in the values:

Percent Yield = (9 g / 21.632 g) × 100% ≈ 41.6%

Therefore, the percent yield of sulfur is approximately 41.6%.