Answer:
the rate at which sulfur dioxide is being produced is 0.8993 kg per second
Explanation:
Given the data in the question;
we write the balanced chemical reaction
2H₂S + 3O₂ → 2SO₂ + 2H₂O
now we calculate the number of moles of oxygen by using an ideal gas equation; PV = nRT
n = PV/RT
so we substitute in our values
n = ( 0.77 × 1.01325 × 10⁵ × 994 × 10⁻³ ) / ( 8.314 × (170 + 273 )
n = 77552.1285 / 3683.102
n = 21.056 mol
so, moles of SO₂ produced = 2/3 × 21.056 = 14.0373 moles
Hence, the production rate will be;
⇒ 14.0373 × 64.066
= 899.3 g/s
= 0.8993 kg/s
Therefore, the rate at which sulfur dioxide is being produced is 0.8993 kg per second