Answer:
1.4 kg/s
Step-by-step explanation:
The reaction of the production of sulfur dioxide is:
H₂S(g) + O₂(g) → SO₂(g) + H₂(g)
By the ideal gas law, the number of moles of oxygen per second is:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (170°C +273 = 443 K).
n = PV/RT
n = (0.77*994)/(0.082*443)
n = 21.07 mol/s
The stoichiometry reaction is 1 mol of O₂:1 mol of SO₂, so the rate of SO₂ is also 21.07 mol/s. The molar mass of SO₂ is:
32 g/mol of S + 2*16 g/mol of O = 64 g/mol
So, the mass rate is the molar mass multiplied by the molar rate:
m = 64 g/mol * 21.07 mol/s
m = 1348.5 g/s
m = 1.4 kg/s