Question

Help please!

Answer 1

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