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Alik [6]
3 years ago
13

S8 + 12 O2 —> 8 SO3 what volume of oxygen would be required to completely react with 45.7g of S8

Chemistry
1 answer:
ahrayia [7]3 years ago
4 0

Answer:

47.9L of oxygen would be required

Explanation:

<em>Assuming oxygen gas is at STP</em>

Based on the reaction:

S₈ + 12 O₂ → 8SO₃

1 mole of S₈ (Molar mass: 256.52g/mol) reacts with 12 moles of oxygen.

To solve this question we must find the moles of S₈ added. With the reaction we can find the moles of O₂ and using PV = nRT we can find the volume of oxygen required:

<em>Moles S₈:</em>

45.7g * (1mol / 256.52g) = 0.178 moles S₈

<em>Moles O₂:</em>

0.178 moles S₈ * (12 moles O₂ / 1mol S₈) = 2.138 moles O₂

<em>Volume O₂:</em>

PV = nRT

V = nRT / P

<em>Where V is volume</em>

<em>n are moles = 2.138 moles</em>

<em>R is gas constant 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K at sTP</em>

<em>P is pressure = 1 atm at STP</em>

<em />

V = 2.138mol*0.082atmL/molK*273.15K / 1atm

V = 47.9L of oxygen would be required

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