Sure, let's solve this step by step.

1. **Write the balanced chemical equation for the oxidation of ammonia:**
$$ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O} $$

2. **Use the Ideal Gas Law to find the moles of dioxygen consumed per second:**
The Ideal Gas Law is given by:
$$ PV = nRT $$
Where:
- $ P $ is the pressure (0.27 atm)
- $ V $ is the volume (876 L)
- $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
- $ T $ is the temperature in Kelvin (193°C + 273.15 = 466.15 K)

Rearrange the Ideal Gas Law to solve for $ n $ (moles of dioxygen):
$$ n = \frac{PV}{RT} $$
$$ n = \frac{(0.27 \text{ atm})(876 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(466.15 \text{ K})} $$
$$ n \approx 6.14 \text{ mol/s} $$

3. **Determine the moles of nitrogen monoxide produced per second:**
From the balanced equation, 5 moles of $ \text{O}_2 $ produce 4 moles of $ \text{NO} $. Therefore, the moles of $ \text{NO} $ produced per second is:
$$ \text{Moles of NO} = \left( \frac{4 \text{ moles NO}}{5 \text{ moles O}_2} \right) \times 6.14 \text{ mol/s} $$
$$ \text{Moles of NO} \approx 4.91 \text{ mol/s} $$

4. **Convert moles of nitrogen monoxide to mass:**
The molar mass of $ \text{NO} $ is:
$$ \text{Molar mass of NO} = 14.01 \text{ g/mol (N)} + 16.00 \text{ g/mol (O)} = 30.01 \text{ g/mol} $$

Therefore, the mass of $ \text{NO} $ produced per second is:
$$ \text{Mass of NO} = 4.91 \text{ mol/s} \times 30.01 \text{ g/mol} $$
$$ \text{Mass of NO} \approx 147.3 \text{ g/s} $$

5. **Convert grams per second to kilograms per second:**
$$ \text{Mass of NO} = 147.3 \text{ g/s} \times \frac{1 \text{ kg}}{1000 \text{ g}} $$
$$ \text{Mass of NO} \approx 0.15 \text{ kg/s} $$

Therefore, the rate at which nitrogen monoxide is being produced is approximately $ 0.15 $ kilograms per second.

Question

Sure, let's solve this step by step.

1. **Write the balanced chemical equation for the oxidation of ammonia:**
   $$ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O} $$

2. **Use the Ideal Gas Law to find the moles of dioxygen consumed per second:**
   The Ideal Gas Law is given by:
   $$ PV = nRT $$
   Where:
   - $ P $ is the pressure (0.27 atm)
   - $ V $ is the volume (876 L)
   - $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
   - $ T $ is the temperature in Kelvin (193°C + 273.15 = 466.15 K)

Rearrange the Ideal Gas Law to solve for $ n $ (moles of dioxygen):
   $$ n = \frac{PV}{RT} $$
   $$ n = \frac{(0.27 \text{ atm})(876 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(466.15 \text{ K})} $$
   $$ n \approx 6.14 \text{ mol/s} $$

3. **Determine the moles of nitrogen monoxide produced per second:**
   From the balanced equation, 5 moles of $ \text{O}_2 $ produce 4 moles of $ \text{NO} $. Therefore, the moles of $ \text{NO} $ produced per second is:
   $$ \text{Moles of NO} = \left( \frac{4 \text{ moles NO}}{5 \text{ moles O}_2} \right) \times 6.14 \text{ mol/s} $$
   $$ \text{Moles of NO} \approx 4.91 \text{ mol/s} $$

4. **Convert moles of nitrogen monoxide to mass:**
   The molar mass of $ \text{NO} $ is:
   $$ \text{Molar mass of NO} = 14.01 \text{ g/mol (N)} + 16.00 \text{ g/mol (O)} = 30.01 \text{ g/mol} $$

Therefore, the mass of $ \text{NO} $ produced per second is:
   $$ \text{Mass of NO} = 4.91 \text{ mol/s} \times 30.01 \text{ g/mol} $$
   $$ \text{Mass of NO} \approx 147.3 \text{ g/s} $$

5. **Convert grams per second to kilograms per second:**
   $$ \text{Mass of NO} = 147.3 \text{ g/s} \times \frac{1 \text{ kg}}{1000 \text{ g}} $$
   $$ \text{Mass of NO} \approx 0.15 \text{ kg/s} $$

Therefore, the rate at which nitrogen monoxide is being produced is approximately $ 0.15 $ kilograms per second.

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