Sulfur trioxide, SO3 , is produced in enormous quantities each year for use in the synthesis of sulfuric acid. S(s) + O2(g) > SO2(g)
2SO2(g) + O2(g) > 2SO3(g)
What volume of O2(g) at 340.°С and a pressure of 4.75 atm is needed to completely convert 6.30 g sulfur to sulfur trioxide?
Volume = L

3.14 L of oxygen (O₂).

Explanation:

We'll begin by calculating the number of mole in 6.3 g of sulphur (S). This can be obtained as follow:

Molar mass of S = 32 g/mol

Mass of S = 6.3 g

Mole of S =.?

Mole = mass / molar mass

Mole of S = 6.3/32

Mole of S = 0.197 mole

Next, we shall write the overall equation of the reaction between sulphur (S) and oxygen (O₂) to produce sulphur trioxide (SO₃) .

This is illustrated below:

S (s) + O₂ (g) —> SO₂ (g)

SO₂ (g) + O₂ (g) —> 2SO₃ (s)

Overall reaction:

2S (s) + 3O₂ (g) —> 2SO₃ (g)

Next, we shall determine the number of mole of oxygen (O₂) needed to completely convert 6.30 g (i.e 0.197 mole) of sulfur.

This is illustrated below:

From the balanced equation above,

2 moles of sulphur (S) required 3 moles of oxygen (O₂) .

Therefore, 0.197 mole of sulphur (S) will require = (0.197 × 3)/2 = 0.296 mole of oxygen (O₂).

Therefore, 0.296 mole of oxygen (O₂) is needed.

Finally, we shall determine the volume of oxygen (O₂) needed as follow:

Number of mole (n) of oxygen (O₂) = 0.296 mole

Temperature (T) = 340 °С = 340 °С + 273 = 613 K

Pressure (P) = 4.75 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of oxygen (O₂) =.?

PV = nRT

4.75 × V = 0.296 × 0.0821 × 613

Divide both side by 4.75

V = (0.296 × 0.0821 × 613) / 4.75

V = 3.14 L

Therefore, 3.14 L of oxygen (O₂) is needed for the reaction.

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.