Question

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: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(\mathrm{S}_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with \(30.0\) grams of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0\) grams of \(\mathrm{O}_{2}\), how many grams of \(\mathrm{S}_{8}\) would be produced, assuming \(98 \%\) yield?

Short answer

The actual yield of S8 produced is 27.65 grams, assuming a 98% yield.

Step-by-step solution

Calculate the moles of reactants

First, we need to find the number of moles for both reactants, H2S and O2. We can use the given masses and their molar masses to do so. Molar mass of H2S = \(1.01 \times 2 + 32.07 = 34.08 \: g/mol\) Moles of H2S = \( \frac{30.0 \: g}{34.08 \: g/mol} = 0.88 \: moles\) Molar mass of O2 = \(16.00 \times 2 = 32.00 \:g/mol\) Moles of O2 = \( \frac{50.0 \: g}{32.00 \: g/mol} = 1.56 \: moles\)

Determine the limiting reactant

Now, we need to find out which reactant is limiting by comparing the mole ratios. For this, divide the moles of each reactant by their respective coefficients in the balanced reaction equation. Mole ratio of H2S = \( \frac{0.88}{8} = 0.11\) Mole ratio of O2 = \( \frac{1.56}{4} = 0.39\) Since the mole ratio of H2S is lower, it is the limiting reactant.

Calculate the theoretical yield of S8

Using the moles of the limiting reactant (H2S) and stoichiometry, we can find the moles of S8 formed in this reaction. Moles of S8 = \( \frac{moles \: of \: H2S}{8} \times 1\) Moles of S8 = \( \frac{0.88}{8} = 0.11 \: moles\) Now, let's find the mass of S8 produced. Molar mass of S8 = \(32.07 \times 8 = 256.56 \: g/mol\) Mass of S8 (theoretical) = moles of S8 × molar mass of S8 Mass of S8 (theoretical) = \(0.11 \: moles \times 256.56 \: g/mol = 28.22 \: g\)

Calculate the actual yield of S8

Since the reaction only has a 98% yield, we need to find out how much S8 is actually produced. Mass of S8 (actual) = \% Yield × Mass of S8 (theoretical) Mass of S8 (actual) = \(0.98 \times 28.22 \: g = 27.65 \: g\) Therefore, 27.65 grams of S8 would be produced, assuming a 98% yield.