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:
(2)
Where:
V: is the volume
: 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:
![n_{S} = \frac{m}{M}](https://tex.z-dn.net/?f=n_%7BS%7D%20%3D%20%5Cfrac%7Bm%7D%7BM%7D)
Where:
: 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
![n_{S}=\frac{m}{M} =\frac{425 g}{32.065 g/mol} = 13.25 moles](https://tex.z-dn.net/?f=n_%7BS%7D%3D%5Cfrac%7Bm%7D%7BM%7D%20%3D%5Cfrac%7B425%20g%7D%7B32.065%20g%2Fmol%7D%20%3D%2013.25%20moles)
From reaction (1) we have that 1 mol of O₂ reacts with 1 mol of S, hence the number of moles of oxygen is:
Finally, the volume of oxygen is (equation (2)):
![V = \frac{13.25 moles*0.082 L*atm/(K*mol)*273 K}{1 atm} = 296.6 L](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B13.25%20moles%2A0.082%20L%2Aatm%2F%28K%2Amol%29%2A273%20K%7D%7B1%20atm%7D%20%3D%20296.6%20L)
Therefore, are necessary 296.6 liters of oxygen for the combustion of sulfur.
I hope it helps you!