Answer:
1.18x10⁸L of SO₂ and 2.36x10⁸L of H₂S
Explanation:
The balanced reaction is:
8SO₂(g) + 16H₂S(g) → 16H₂O(l) + 3S₈(s)
To solve this question we must find the moles of S₈ in 4.50x10⁵kg. With these moles and the reaction we can find the moles of SO₂ needed to react (Twice these moles = Moles Of H₂S needed). Using PV = nRT we can find the volume of the gas required:
<em>Moles S₈ - molar mass: 256.52g/mol-</em>
4.50x10⁵kg = 4.50x10⁸g * (1mol / 256.52g) =
1.75x10⁶ moles S₈
<em>Moles SO₂:</em>
1.75x10⁶ moles S₈ * (8mol SO₂ / 3mol S₈) = 4.68x10⁶ moles SO₂
<em>Moles H₂S:</em>
4.68x10⁶ moles SO₂ * 2 = 9.36x10⁶ moles H₂S
The volume could be obtained as follows:
PV = nRT
V = nRT / P
<em>V is volume in liters</em>
<em>n are moles: 4.68x10⁶ moles SO₂ and 9.36x10⁶ moles H₂S</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is absolute temperature = 22°C + 273.15 = 295.15K</em>
<em>P is pressure = 0.961atm</em>
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Replacing:
Volume SO₂ and H₂S:
4.68x10⁶ moles * 0.082atmL/molK * 295.15K / 0.961atm =
<h3>1.18x10⁸L of SO₂ and:</h3>
<em>9.36x10⁶ moles H₂S</em> * 0.082atmL/molK * 295.15K / 0.961atm =
<h3>2.36x10⁸L of H₂S</h3>
