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Determine the volume of H₂S (at 375 K and 1.20 atm) needed to produce 55.0 g of S. Assume that there is excess SO₂ present.2 H₂S(g) + SO₂(g) ? 3 S(s) + 2 H₂O(g)

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Final answer:

To find the volume of H₂S required to produce 55.0 g of S, calculate the moles of S, use the mole ratio from the balanced chemical equation to find moles of H₂S, and then apply the ideal gas law to find the volume of H₂S needed, which is approximately 35.6 liters.

Step-by-step explanation:

The student is asking how to calculate the volume of hydrogen sulfide (H₂S) gas needed to produce 55.0 g of sulfur (S) given that the gas is at a temperature of 375 K and a pressure of 1.20 atm, with an excess of sulfur dioxide (SO₂) present. The balanced chemical equation for the reaction is:

2 H₂S(g) + SO₂(g) → 3 S(s) + 2 H₂O(g)

First, we calculate the moles of sulfur by using its molar mass:

55.0 g S × (1 mol S / 32.065 g S) = 1.715 mol S

From the stoichiometry of the balanced equation, 2 moles of H₂S are required for every 3 moles of S produced. Now we can find the moles of H₂S needed:

1.715 mol S × (2 mol H₂S / 3 mol S) = 1.143 mol H₂S

Next, we use the ideal gas law to calculate the volume of H₂S:

V = (nRT) / P

V = (1.143 mol × (0.0821 L·atm/mol·K) × 375 K) / 1.20 atm

V ≈ 35.6 L of H₂S

Therefore, approximately 35.6 liters of H₂S is required to produce 55.0 g of S under the given conditions.

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