Answer:
V = 16.3 L
Explanation:
Given data:
Volume of oxygen needed = ?
Temperature = standard = 273 K
Pressure = standard = 1 atm
Volume of sulfur trioxide = 32.6 L
Solution:
Chemical equation:
2SO₂ + O₂ → 2SO₃
Number of moles of SO₃:
PV = nRT
1 atm × 32.6 L = n×0.0821 atm.L/mol.K ×273 K
32.6 atm.L =n× 22.41 atm.L/mol
n = 32.6 atm.L / 22.41 atm.L/mol
n = 1.45 mol
now we will compare the moles of oxygen with sulfur trioxide.
SO₃ : O₂
2 : 1
1.45 : 1/2×1.45 = 0.725 mol
Volume of oxygen:
PV = nRT
1 atm × V = 0.725 mol×0.0821 atm.L/mol.K ×273 K
V = 16.25 atm.L /1 atm
V = 16.3 L
