Answer:
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.
