In the reaction below, 22 g of H2S with excess
O2 produced 10 g of sulfur.
?H2S + ?O2 ! ? S + ?H2O.
What is the percent yield of sulfur?
Answer in units of %
Balance the equation.
Convert 22 g H2S to mols. mols = grams/molar mass
Using the coefficients in the balanced equation, convert mols H2S to mols S.
Finally, convert mols S to grams S. g = mols x atomic mass. This is the theoretical yield (TY). Actual yield (AY) is 10 g from the problem.
%S = (AY/TY)*100 = ?
i got 8.71%
I don't think so. If you will post your work I will find the error for you.
To find the percent yield of sulfur, we need to compare the actual yield (10 g) with the theoretical yield. The theoretical yield is the amount of sulfur that would be obtained if the reaction had gone to completion.
First, let's balance the chemical equation:
2H2S + 3O2 -> 2S + 2H2O.
From the equation, we can see that 2 moles of H2S produce 2 moles of S. Therefore, the molar ratio between H2S and S is 1:1.
Now, let's calculate the moles of H2S:
molar mass of H2S = 2(1.0078) + 32.06 = 34.076 g/mol
moles of H2S = mass of H2S / molar mass of H2S = 22 g / 34.076 g/mol ≈ 0.646 moles
Since the molar ratio between H2S and S is 1:1, the moles of S produced will also be 0.646 moles.
Now, we can calculate the theoretical yield of sulfur:
mass of S = moles of S * molar mass of S = 0.646 moles * 32.06 g/mol ≈ 20.7 g
Now that we have the actual yield (10 g) and the theoretical yield (20.7 g), we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100
= (10 g / 20.7 g) * 100
≈ 48.3 %
Therefore, the percent yield of sulfur in this reaction is approximately 48.3%.