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
