In the reaction below, 16 g of H2S with excess O2 produced 8 g of sulfur.
? H2S + ? O2 → ? S + ? H2O .
What is the percent yield of sulfur?
Answer in units of %
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2H2S + O2 ==> S + 2H2O
mols H2S = grams/molar mass
Use the coefficients in the balanced equation to convert mols H2S to mols S. (hint: so mols S = 1/2 mols H2S.)
Convert mols S to grams S. grams = mols x molar mass. This is the theoretical yield (TY). The actual yield(AY) is 8.0g from the problem.
%yield = (AY/TY)*100 = ?
I still don't understand how to get the theroetical yield.
I gave you a step by step procedure.
mols H2S = grams H2S/molar mass H2S = 16/34 = about 0.47 mols H2S.
Now convert mols H2S to mols S. In the balanced equation you have 2 mols H2S forming 2 mols S (yes, I made a typo. The balanced equation is
2H2S + O2 ==> 2S + 2H2O so 0.47 mols H2S = 0.47 mols S.
Convert to grams S or g = mols x atomic mass S = 0.47 x 32 = about 15 and this is the theoretical yield which is about twice if you followed the wrong equation I posted first.
%yield = (8/15) = about 53%
You can go through and clean up my estimates here and there.
Well, let's do some math and see what we end up with.
We start with 16 g of H2S, and the molar mass of H2S is about 34 g/mol. So, we have:
16 g H2S * (1 mol H2S / 34 g H2S) = about 0.47 mol H2S
From the balanced equation, we can see that for every 1 mol of H2S, we should get 1 mol of S. So, we should expect to get:
0.47 mol S
The molar mass of Sulfur is about 32 g/mol. So, the expected mass of Sulfur is:
0.47 mol S * (32 g S / 1 mol S) = about 15 g of Sulfur
But the actual yield is given as 8 g of Sulfur. So, the percent yield can be calculated using this formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Using the values we've calculated:
Percent Yield = (8 g S / 15 g S) * 100 = about 53.33%
So, the percent yield of Sulfur in this reaction is approximately 53.33%. Well, at least it's more than half full!
To calculate the percent yield of sulfur, we need to compare the actual yield with the theoretical yield. The actual yield is given in the problem statement as 8 g of sulfur.
To find the theoretical yield, we need to determine the stoichiometry of the reaction, which represents the balanced ratio of reactants and products.
From the chemical equation given:
? H2S + ? O2 → ? S + ? H2O
We can determine the stoichiometry by comparing the coefficients of the balanced equation. In this case, we see that the coefficient of H2S is 1, meaning that for every 1 mole of H2S, we produce 1 mole of S.
Since we know the mass of H2S (16 g), we can convert it to moles using the molar mass of H2S, which is 34.08 g/mol (the molar mass of hydrogen is 1.01 g/mol, and the molar mass of sulfur is 32.06 g/mol):
16 g H2S * (1 mol H2S / 34.08 g H2S) = 0.469 mol H2S
Since the stoichiometric ratio is 1:1 between H2S and S, the theoretical yield of sulfur is also 0.469 mol.
Now, we can calculate the percent yield using the formula:
Percent Yield = (actual yield / theoretical yield) * 100
Plugging in the actual yield (8 g) and the theoretical yield (0.469 mol):
Percent Yield = (8 g / (0.469 mol * 32.06 g/mol)) * 100
Simplifying that:
Percent Yield = (8 g / 15.13914 g) * 100
Percent Yield = 52.92%
Therefore, the percent yield of sulfur in this reaction is 52.92%.