Final answer:
The volume of H₂S gas required to produce 55.0 g of sulfur at 375 K and 1.20 atm pressure is approximately 43.12 liters. This is calculated using the molar mass of sulfur and the ideal gas law.
Step-by-step explanation:
To determine the volume of H₂S gas needed to produce 55.0 g of sulfur (S), we first need to use the chemical equation:
H₂S(g) + SO₂(g) → H₂O(l) + S(s)
This reaction is already balanced with a 1:1 molar ratio between H₂S and solid sulfur. To find the moles of sulfur produced, we use the molar mass of sulfur (approximately 32.07 g/mol).
Moles of S = mass of S / molar mass of S
Moles of S = 55.0 g / 32.07 g/mol ≈ 1.715 mol
Because the molar ratio between H₂S and S is 1:1, the moles of H₂S needed will also be 1.715 mol. To find the volume of H₂S at the given conditions (375 K and 1.20 atm), we use the ideal gas law:
PV = nRT
Where P is pressure, V is volume, n is moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin. We rearrange to solve for V:
V = nRT / P
V = (1.715 mol) × (0.0821 L·atm/mol·K) × (375 K) / (1.20 atm)
V ≈ 43.12 L
Thus, about 43.12 liters of H₂S gas are needed to produce 55.0 g of sulfur under the given conditions.