What mass of sulfur has to burn to produce 4.5l so2 at 300°c and 101 kpa in the following reaction? s(s)+o2(g) so,(g) a. 13.5 g s b. 3.07 g s c. 68.8 g s d. 41.0 g s

option B. 3.07 g

Explanation:

1) given reaction:

S(s) + O₂ (g) → SO(g)

2) Balanced chemical equation:

2S(s) + O₂ (g) → 2SO(g)

3) Theoretical mole ratios:

2 mol S : 1 mol O₂ : 2 mol SO

3) number of moles of 4.5 liter SO₂ at 300°C and 101 kPa

use the ideal gas equation:

pV = nRT

with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)

=> n = pV / (RT) =

n =  [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K)  × 573.15 K ]

n = 0.0954 mol SO

4) proportion with the theoretical ratio S / SO

 2 mol S                   x
=
 2 mol SO      0.0954 mol SO

=> x = 0.0954 mol S.

5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol

mass = number of moles × atomic mass

mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S

6) Therefore the answer is the option B. 3.07 g

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