To solve this problem, we need to follow these steps:

1. **Write the balanced chemical equation for the Claus process:**
$$ 2 H_2S + O_2 \rightarrow 2 S + 2 H_2O $$

2. **Determine the molar volume of dioxygen gas at the given conditions using the Ideal Gas Law:**
$$ PV = nRT $$
Where:
- $ P $ is the pressure (0.52 atm)
- $ V $ is the volume (250 L/s)
- $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
- $ T $ is the temperature in Kelvin (290°C + 273 = 563 K)

Rearrange the Ideal Gas Law to solve for $ n $ (moles of $ O_2 $):
$$ n = \frac{PV}{RT} $$
$$ n = \frac{(0.52 \text{ atm})(250 \text{ L/s})}{(0.0821 \text{ L·atm/(K·mol)})(563 \text{ K})} $$
$$ n \approx 2.80 \text{ mol/s} $$

3. **Use stoichiometry to find the rate of sulfur dioxide production:**
From the balanced equation, 1 mole of $ O_2 $ produces 1 mole of $ SO_2 $. Therefore, the rate of $ SO_2 $ production is the same as the rate of $ O_2 $ consumption:
$$ \text{Rate of } SO_2 = 2.80 \text{ mol/s} $$

4. **Convert moles of $ SO_2 $ to mass:**
The molar mass of $ SO_2 $ is:
$$ 32.07 \text{ (S)} + 2 \times 16.00 \text{ (O)} = 64.07 \text{ g/mol} $$

Therefore, the mass of $ SO_2 $ produced per second is:
$$ \text{Mass of } SO_2 = 2.80 \text{ mol/s} \times 64.07 \text{ g/mol} $$
$$ \text{Mass of } SO_2 \approx 179.40 \text{ g/s} $$

5. **Convert grams per second to kilograms per second:**
$$ 179.40 \text{ g/s} = 0.1794 \text{ kg/s} $$

6. **Round to 2 significant digits:**
$$ \text{Rate of } SO_2 \text{ production} \approx 0.18 \text{ kg/s} $$

So, the rate at which sulfur dioxide is being produced is approximately 0.18 kilograms per second.

Question

To solve this problem, we need to follow these steps:

1. **Write the balanced chemical equation for the Claus process:**
   $$ 2 H_2S + O_2 \rightarrow 2 S + 2 H_2O $$

2. **Determine the molar volume of dioxygen gas at the given conditions using the Ideal Gas Law:**
   $$ PV = nRT $$
   Where:
   - $ P $ is the pressure (0.52 atm)
   - $ V $ is the volume (250 L/s)
   - $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
   - $ T $ is the temperature in Kelvin (290°C + 273 = 563 K)

Rearrange the Ideal Gas Law to solve for $ n $ (moles of $ O_2 $):
   $$ n = \frac{PV}{RT} $$
   $$ n = \frac{(0.52 \text{ atm})(250 \text{ L/s})}{(0.0821 \text{ L·atm/(K·mol)})(563 \text{ K})} $$
   $$ n \approx 2.80 \text{ mol/s} $$

3. **Use stoichiometry to find the rate of sulfur dioxide production:**
   From the balanced equation, 1 mole of $ O_2 $ produces 1 mole of $ SO_2 $. Therefore, the rate of $ SO_2 $ production is the same as the rate of $ O_2 $ consumption:
   $$ \text{Rate of } SO_2 = 2.80 \text{ mol/s} $$

4. **Convert moles of $ SO_2 $ to mass:**
   The molar mass of $ SO_2 $ is:
   $$ 32.07 \text{ (S)} + 2 \times 16.00 \text{ (O)} = 64.07 \text{ g/mol} $$

Therefore, the mass of $ SO_2 $ produced per second is:
   $$ \text{Mass of } SO_2 = 2.80 \text{ mol/s} \times 64.07 \text{ g/mol} $$
   $$ \text{Mass of } SO_2 \approx 179.40 \text{ g/s} $$

5. **Convert grams per second to kilograms per second:**
   $$ 179.40 \text{ g/s} = 0.1794 \text{ kg/s} $$

6. **Round to 2 significant digits:**
   $$ \text{Rate of } SO_2 \text{ production} \approx 0.18 \text{ kg/s} $$

So, the rate at which sulfur dioxide is being produced is approximately 0.18 kilograms per second.

Knowee AI · Accepted Answer