Answer:
V = 296.6 liters of oxygen.
Explanation:
The reaction of combustion of sulfur is:
O₂(g) + S(s) → SO₂(g) (1)
To find the volume of the oxygen we need to use the Ideal Gas Law:
[tex] V = \frac{n_{O}RT}{P} [/tex] (2)
Where:
V: is the volume
[tex]n_{O}[/tex]: is the number of moles of oxygen
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 273 K (at STP)
P: is the pressure = 1 atm (at STP)
So we need to find the number of moles of oxygen that reacts with sulfur:
[tex]n_{S} = \frac{m}{M}[/tex]
Where:
[tex]n_{S}[/tex]: is the number of moles of sulfur
m: is the mass of sulfur = 425 g
M: is the molar mass of sulfur = 32.065 g/mol
[tex]n_{S}=\frac{m}{M} =\frac{425 g}{32.065 g/mol} = 13.25 moles[/tex]
From reaction (1) we have that 1 mol of O₂ reacts with 1 mol of S, hence the number of moles of oxygen is:
[tex] n_{S} = n_{O} = 13.25 moles [/tex]
Finally, the volume of oxygen is (equation (2)):
[tex]V = \frac{13.25 moles*0.082 L*atm/(K*mol)*273 K}{1 atm} = 296.6 L[/tex]
Therefore, are necessary 296.6 liters of oxygen for the combustion of sulfur.
I hope it helps you!
