Question

The principal source of sulfur on earth is deposits of free sulfur occurring mainly in volcanically Active regions. The offer was initially formed by the reaction between the two volcanic vapors S02 and H2S to form H2O(I) and S8(s). What volume of each gas, at 0.961 atm and 22°C, what is needed to form a sulfur deposit of 4.50×10 to the power of 5 kg On the slopes of a volcano in Hawaii?

Answer 1

1.18x10⁸L of SO₂ and 2.36x10⁸L of H₂S

Step-by-step 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:

Moles S₈ - molar mass: 256.52g/mol-

4.50x10⁵kg = 4.50x10⁸g * (1mol / 256.52g) =

1.75x10⁶ moles S₈

Moles SO₂:

1.75x10⁶ moles S₈ * (8mol SO₂ / 3mol S₈) = 4.68x10⁶ moles SO₂

Moles H₂S:

4.68x10⁶ moles SO₂ * 2 = 9.36x10⁶ moles H₂S

The volume could be obtained as follows:

PV = nRT

V = nRT / P

V is volume in liters

n are moles: 4.68x10⁶ moles SO₂ and 9.36x10⁶ moles H₂S

R is gas constant = 0.082atmL/molK

T is absolute temperature = 22°C + 273.15 = 295.15K

P is pressure = 0.961atm

Replacing:

Volume SO₂ and H₂S:

4.68x10⁶ moles * 0.082atmL/molK * 295.15K / 0.961atm =

1.18x10⁸L of SO₂ and:

9.36x10⁶ moles H₂S * 0.082atmL/molK * 295.15K / 0.961atm =

2.36x10⁸L of H₂S