Answer:
The rate at which sulfur dioxide is being produced is 0.90 kg/s.
Explanation:
Volume of oxygen gas consumed in second ,V= 994 L
Pressure of the gas = p
Temperature of the gas = T = 170°C= 170 + 273 K=443 K
Moles of oxygen gas consumed in a second = n
[tex]PV=nRT[/tex] ( ideapl gas equation)
[tex]n=\frac{PV}{RT}=\frac{0.77 atm\times 994 L}{0.0821 atm L/mol K\times 443 K}[/tex]
n = 21.044 mole
Moles of dioxygen gas consumed per second = 21.044 mol
[tex]2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)[/tex] (Claus process)
According to reaction, 3 moles of dioxygen gives 2 moles of sulfur dioxide gas.Then 21.044 moles of dioxygen will give;
[tex]\frac{2}{3}\times 21.044 ol=14.029 mol[/tex] of sulfur dioxide
Mass of 14.029 moles of sulfur dioxide gas;
14.029 mol × 64 g/mol = 897.86 g
897.86 g = 0.89786 kg ≈ 0.90 kg
Mass of sulfur dioxide produced per second = 0.90 kg
The rate at which sulfur dioxide is being produced is 0.90 kg/s.
