Answer :
Answer:
[tex]0.21\ \text{kg/s}[/tex]
Explanation:
P = Pressure = [tex]0.48\ \text{atm}=0.48\times 101325\ \text{Pa}[/tex]
V = Volume = [tex]450\ \text{L/s}=450\times 10^{-3}\ \text{m}^3/\text{s}[/tex]
R = Gas constant = [tex]8.314\ \text{J/mol K}[/tex]
T = Temperature = [tex](264+273.15)\ \text{K}[/tex]
The reaction is
[tex]2H_2S+3O_2\rightarrow 2SO_2+2H_2O[/tex]
From ideal gas equation we have
[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{0.48\times101325\times 450\times 10^{-3}}{8.314\times (264+273.15)}\\\Rightarrow n=4.9\ \text{mol}[/tex]
Moles of [tex]SO_2[/tex] produced is
[tex]\dfrac{2}{3}\times 4.9=3.267\ \text{moles}[/tex]
Molar mass of [tex]SO_2[/tex] = 64.066 g/mol
Production rate is
[tex]3.267\times 64.066=209.3\ \text{g/s}=0.21\ \text{kg/s}[/tex]
The rate at which sulfur dioxide is being produced [tex]0.21\ \text{kg/s}[/tex].