In the oxydation of sulfur from volcanic activities, how many liters of sulfur is processed if 255 g of oxygen from the air is consumed at 283 K and 4 atm pressure forming sulfur dioxide? * with solution pls

46.2 L

Explanation:

The equation of the reaction between sulphur and oxygen;

S(s) + O2(g) > SO2(g)

Number of moles of oxygen reacted = 255g/32 g/mol = 7.97 moles

Now;

If 1 mole of oxygen reacts with 1 mole of sulphur

7.97 moles of oxygen will also react with 7.97 moles of sulphur

From the ideal gas equation;

PV = nRT

P = 4 atm

T = 283 K

n= 7.97

V= ?

R= 0.082 atm LK-1mol-1

V = nRT/P

V= 7.97 * 0.082 * 283/4

V = 46.2 L

i would say b

explanation:

but i think i would need more information than that