In the reaction below, 17 g of H2S with excess
O2 produced 10 g of sulfur.
2 H2S + O2 → 2 S + 2 H2O .
What is the percent yield of sulfur?
Answer in units of %.
To calculate the percent yield of sulfur in this reaction, we need to first determine the theoretical yield of sulfur and then compare it to the actual yield.
1) Calculate the molar mass of H2S:
- Hydrogen (H) has a molar mass of 1 g/mol.
- Sulfur (S) has a molar mass of 32.1 g/mol.
So, the molar mass of H2S is (2*1) + 32.1 = 34.1 g/mol.
2) Convert the mass of H2S to moles:
Given that the mass of H2S is 17 g, we can calculate the number of moles using the formula:
moles = mass / molar mass.
moles = 17 g / 34.1 g/mol = 0.498 mol.
3) Use stoichiometry to determine the theoretical yield of sulfur:
From the balanced equation, we can see that 2 moles of H2S produce 2 moles of sulfur.
So, the moles of sulfur produced would also be 0.498 mol.
4) Calculate the mass of sulfur produced for the theoretical yield:
To find the mass of sulfur, multiply the moles of sulfur by the molar mass of sulfur:
mass = moles * molar mass.
mass = 0.498 mol * 32.1 g/mol = 15.99 g.
5) Calculate the percent yield:
The percent yield is given by the formula:
percent yield = (actual yield / theoretical yield) * 100.
Given that the actual yield is 10 g and the theoretical yield is 15.99 g, we can calculate:
percent yield = (10 g / 15.99 g) * 100 = 62.5%.
Therefore, the percent yield of sulfur in this reaction is 62.5%.
To calculate the percent yield of sulfur in this reaction, you need to compare the actual yield (the amount of sulfur produced) to the theoretical yield (the amount of sulfur that would be produced if the reaction went to completion).
First, you need to determine the number of moles of sulfur produced. To do this, convert the mass of sulfur produced to moles.
Given:
Mass of sulfur produced = 10 g
The molar mass of sulfur (S) is 32.06 g/mol. Therefore, divide the mass of sulfur by the molar mass to get the number of moles of sulfur produced:
Number of moles of sulfur = 10 g / 32.06 g/mol = 0.312 mol
Next, you need to find the number of moles of sulfur that would be produced if the reaction went to completion. This can be determined by stoichiometry, using the balanced equation:
2 H2S + O2 → 2 S + 2 H2O
From the balanced equation, you can see that for every 2 moles of H2S reacted, 2 moles of sulfur are produced. Therefore, you need to calculate the number of moles of H2S used in the reaction.
Given:
Mass of H2S used = 17 g
The molar mass of H2S is 34.08 g/mol. Therefore, divide the mass of H2S by the molar mass to get the number of moles of H2S used:
Number of moles of H2S used = 17 g / 34.08 g/mol = 0.499 mol
Since the stoichiometric ratio between H2S and S is 2:2, the number of moles of sulfur that would be produced if the reaction went to completion is also 0.499 mol.
Now that you have the actual yield (0.312 mol) and the theoretical yield (0.499 mol), you can calculate the percent yield.
Percent yield = (actual yield / theoretical yield) × 100%
Percent yield = (0.312 mol / 0.499 mol) × 100%
Percent yield = 0.625 × 100%
Percent yield = 62.5%
Therefore, the percent yield of sulfur in this reaction is 62.5%.