Answer:
![0.21\ \text{kg/s}](https://img.qammunity.org/2022/formulas/chemistry/college/chohjch2pgev0vnyc3zy8pfjajkm343uap.png)
Step-by-step explanation:
P = Pressure =
![0.48\ \text{atm}=0.48* 101325\ \text{Pa}](https://img.qammunity.org/2022/formulas/chemistry/college/kuys97jqotdb6omfrhv1r85gpd0t2kycmo.png)
V = Volume =
![450\ \text{L/s}=450* 10^(-3)\ \text{m}^3/\text{s}](https://img.qammunity.org/2022/formulas/chemistry/college/9y893csutg8yozv01d26bj5dfya06bp17h.png)
R = Gas constant =
![8.314\ \text{J/mol K}](https://img.qammunity.org/2022/formulas/chemistry/college/rn5oz8d33ne1x5a482jalg3qepbno1vv1t.png)
T = Temperature =
![(264+273.15)\ \text{K}](https://img.qammunity.org/2022/formulas/chemistry/college/ss3rpc82axjztgzepvf1ju03jpexunru9h.png)
The reaction is
![2H_2S+3O_2\rightarrow 2SO_2+2H_2O](https://img.qammunity.org/2022/formulas/chemistry/college/dydybnjecx1owjt7f13fxtizq229oss9pt.png)
From ideal gas equation we have
![PV=nRT\\\Rightarrow n=(PV)/(RT)\\\Rightarrow n=(0.48*101325* 450* 10^(-3))/(8.314* (264+273.15))\\\Rightarrow n=4.9\ \text{mol}](https://img.qammunity.org/2022/formulas/chemistry/college/u3fewa1zsw7lztx2u4urn6mrq35ex9exg8.png)
Moles of
produced is
![(2)/(3)* 4.9=3.267\ \text{moles}](https://img.qammunity.org/2022/formulas/chemistry/college/epgbdyobsyf73c8z13pwtrz6fcv7u0t3by.png)
Molar mass of
= 64.066 g/mol
Production rate is
![3.267* 64.066=209.3\ \text{g/s}=0.21\ \text{kg/s}](https://img.qammunity.org/2022/formulas/chemistry/college/a2wb54wh8hrqc761d2s8bdvz6tqpcwebfg.png)
The rate at which sulfur dioxide is being produced
.