Answer:
47.9L of oxygen would be required
Step-by-step explanation:
Assuming oxygen gas is at STP
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:
Moles S₈:
45.7g * (1mol / 256.52g) = 0.178 moles S₈
Moles O₂:
0.178 moles S₈ * (12 moles O₂ / 1mol S₈) = 2.138 moles O₂
Volume O₂:
PV = nRT
V = nRT / P
Where V is volume
n are moles = 2.138 moles
R is gas constant 0.082atmL/molK
T is absolute temperature = 273.15K at sTP
P is pressure = 1 atm at STP
V = 2.138mol*0.082atmL/molK*273.15K / 1atm
V = 47.9L of oxygen would be required