Hydrogen sulfide is an impurity in natural gas that must be removed.
One common removal method is called the Claus
process, which relies on the reaction:
8 H2S(g) + 4 O2(g)--> S(l) + 8 H2O(g)
Under optimal conditions the Claus process gives 98% yield of
S8 from H2S. If you started with 30.0 grams of H2S and 50.0
grams of O2, how many grams of S8 would be produced, assuming
98% yield?
First I would balance the equation by making it 8S. I'm sure this is just a typo.
Since amounts are given for BOTH reactants we know this is a limiting reagent problem. I solve this by working two stoichiometry problems, the first one with 30.0g H2S and all the O2 needed; the second with 50.0 g O2 and all the H2S needed. You will get two answers for the moles S, of course, and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is a worked example of how to do the stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck.
To find the number of grams of S8 produced, we need to calculate the number of moles of H2S and O2, determine the limiting reactant, and then use stoichiometry to find the moles of S8 produced. Finally, we can convert the moles of S8 to grams.
1. Calculate the number of moles of H2S:
- The molar mass of H2S is 32.06 g/mol (2 * 1.008 g/mol for hydrogen + 32.06 g/mol for sulfur).
- Moles of H2S = mass of H2S / molar mass of H2S
= 30.0 g / 32.06 g/mol
≈ 0.936 mol
2. Calculate the number of moles of O2:
- The molar mass of O2 is 32.00 g/mol (16.00 g/mol for oxygen * 2).
- Moles of O2 = mass of O2 / molar mass of O2
= 50.0 g / 32.00 g/mol
≈ 1.5625 mol
3. Determine the limiting reactant:
To find the limiting reactant, we compare the mole ratios of H2S and O2 to the balanced equation. The stoichiometric ratio from the balanced equation is 8:4, which simplifies to 2:1.
- Moles of H2S / 2 = Moles of O2 / 1
- 0.936 mol / 2 = 1.5625 mol / 1
- 0.468 = 1.5625
Since 0.468 is smaller, H2S is the limiting reactant.
4. Find the moles of S8 produced:
From the balanced equation, we know that for every 8 moles of H2S reacted, we get 1 mole of S8.
- Using the stoichiometric ratio, moles of S8 = Moles of H2S / 8
= 0.936 mol / 8
≈ 0.117 mol
5. Calculate the mass of S8 produced:
- The molar mass of S8 is 256.52 g/mol (8 * 32.06 g/mol for sulfur).
- Mass of S8 = moles of S8 * molar mass of S8
= 0.117 mol * 256.52 g/mol
≈ 30.00184 g
Therefore, approximately 30.0 grams of S8 would be produced, assuming a 98% yield.
To find the number of grams of S8 produced, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the amount of S8 that can be produced from both H2S and O2.
1. H2S:
Given mass of H2S = 30.0 grams
Molar mass of H2S = 34.08 g/mol
To find the moles of H2S:
moles of H2S = mass / molar mass = 30.0 g / 34.08 g/mol ≈ 0.88 mol
From the balanced equation, the stoichiometry states that 8 moles of H2S produce 1 mol of S8:
moles of S8 from H2S = (0.88 mol H2S) × (1 mol S8 / 8 mol H2S) = 0.11 mol S8
2. O2:
Given mass of O2 = 50.0 grams
Molar mass of O2 = 32.00 g/mol
To find the moles of O2:
moles of O2 = mass / molar mass = 50.0 g / 32.00 g/mol ≈ 1.56 mol
From the balanced equation, the stoichiometry states that 4 moles of O2 produce 1 mol of S8:
moles of S8 from O2 = (1.56 mol O2) × (1 mol S8 / 4 mol O2) = 0.39 mol S8
Since the Clauss process gives a 98% yield of S8, we need to determine the actual yield:
Actual yield = 98% × Theoretical yield
Theoretical yield of S8 would be the smaller value obtained from the two calculations above:
Theoretical yield = min(moles of S8 from H2S, moles of S8 from O2)
= min(0.11 mol S8, 0.39 mol S8) = 0.11 mol S8
Now let's calculate the mass of S8 produced:
Mass of S8 = moles of S8 × molar mass of S8
= 0.11 mol × 256.48 g/mol ≈ 28.2 grams
Therefore, assuming a 98% yield, approximately 28.2 grams of S8 would be produced.