Question

How many liters of oxygen are necessary for the combustion of 425 g of sulfur, assuming that the reaction occurs at STP

Answer 1

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:

$V = \frac{n_{O}RT}{P}$   (2)

Where:

V: is the volume

$n_{O}$: 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:

$n_{S} = \frac{m}{M}$

Where:

$n_{S}$: 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                                                                            

$n_{S}=\frac{m}{M} =\frac{425 g}{32.065 g/mol} = 13.25 moles$

From reaction (1) we have that 1 mol of O₂ reacts with 1 mol of S, hence the number of moles of oxygen is:

$n_{S} = n_{O} = 13.25 moles$  

Finally, the volume of oxygen is (equation (2)):

$V = \frac{13.25 moles*0.082 L*atm/(K*mol)*273 K}{1 atm} = 296.6 L$

Therefore, are necessary 296.6 liters of oxygen for the combustion of sulfur.

             

I hope it helps you!