Answer :
To solve the problem of determining how much NO is formed from the reaction of ammonia (NH₃) and oxygen (O₂), we need to follow these steps:
1. Determine Molar Masses:
- Molar mass of NH₃ (Ammonia) = 17.03 g/mol
- Molar mass of O₂ (Oxygen) = 32.00 g/mol
- Molar mass of NO (Nitric Oxide) = 30.01 g/mol
2. Convert Mass to Moles:
- Find the number of moles of NH₃:
[tex]\[
\text{Moles of NH₃} = \frac{\text{mass of NH₃}}{\text{molar mass of NH₃}} = \frac{55,000 \text{ g}}{17.03 \text{ g/mol}} \approx 3,229.59 \text{ mol}
\][/tex]
- Find the number of moles of O₂:
[tex]\[
\text{Moles of O₂} = \frac{\text{mass of O₂}}{\text{molar mass of O₂}} = \frac{139,000 \text{ g}}{32.00 \text{ g/mol}} \approx 4,343.75 \text{ mol}
\][/tex]
3. Determine the Limiting Reactant:
- According to the balanced chemical equation, 4 moles of NH₃ react with 5 moles of O₂ to produce 4 moles of NO.
- Calculate the ratio of moles:
- For NH₃:
[tex]\[
\text{Ratio} = \frac{\text{Moles of NH₃}}{4} \approx 0.807
\][/tex]
- For O₂:
[tex]\[
\text{Ratio} = \frac{\text{Moles of O₂}}{5} \approx 0.869
\][/tex]
- The limiting reactant is NH₃ since it has the smaller ratio.
4. Calculate Moles of NO Produced:
- Use the limiting reactant's ratio to calculate the moles of NO:
[tex]\[
\text{Moles of NO} = \text{Limiting Ratio} \times 4 \approx 3,229.59 \text{ mol}
\][/tex]
5. Convert Moles of NO to Mass:
- Calculate the mass of NO produced:
[tex]\[
\text{Mass of NO} = \text{Moles of NO} \times \text{molar mass of NO} = 3,229.59 \, \text{mol} \times 30.01 \, \text{g/mol} \approx 96,920.14 \, \text{g}
\][/tex]
Thus, the mass of NO formed is approximately 96.92 kg.
1. Determine Molar Masses:
- Molar mass of NH₃ (Ammonia) = 17.03 g/mol
- Molar mass of O₂ (Oxygen) = 32.00 g/mol
- Molar mass of NO (Nitric Oxide) = 30.01 g/mol
2. Convert Mass to Moles:
- Find the number of moles of NH₃:
[tex]\[
\text{Moles of NH₃} = \frac{\text{mass of NH₃}}{\text{molar mass of NH₃}} = \frac{55,000 \text{ g}}{17.03 \text{ g/mol}} \approx 3,229.59 \text{ mol}
\][/tex]
- Find the number of moles of O₂:
[tex]\[
\text{Moles of O₂} = \frac{\text{mass of O₂}}{\text{molar mass of O₂}} = \frac{139,000 \text{ g}}{32.00 \text{ g/mol}} \approx 4,343.75 \text{ mol}
\][/tex]
3. Determine the Limiting Reactant:
- According to the balanced chemical equation, 4 moles of NH₃ react with 5 moles of O₂ to produce 4 moles of NO.
- Calculate the ratio of moles:
- For NH₃:
[tex]\[
\text{Ratio} = \frac{\text{Moles of NH₃}}{4} \approx 0.807
\][/tex]
- For O₂:
[tex]\[
\text{Ratio} = \frac{\text{Moles of O₂}}{5} \approx 0.869
\][/tex]
- The limiting reactant is NH₃ since it has the smaller ratio.
4. Calculate Moles of NO Produced:
- Use the limiting reactant's ratio to calculate the moles of NO:
[tex]\[
\text{Moles of NO} = \text{Limiting Ratio} \times 4 \approx 3,229.59 \text{ mol}
\][/tex]
5. Convert Moles of NO to Mass:
- Calculate the mass of NO produced:
[tex]\[
\text{Mass of NO} = \text{Moles of NO} \times \text{molar mass of NO} = 3,229.59 \, \text{mol} \times 30.01 \, \text{g/mol} \approx 96,920.14 \, \text{g}
\][/tex]
Thus, the mass of NO formed is approximately 96.92 kg.
