Question

Help please!

Assuming S.T.P., how many liters of oxygen when combined with excess sulfur dioxide are needed to produce 32.6 L of sulfur trioxide?

A) 0.728 L
B) 16.3 L
C) 32.6 L
D) 65.2 L

Answer 1

V = 16.3 L

Step-by-step 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