To determine the limiting reactant when 4.0 grams of NH₃ reacts with 8.0 grams of O₂ according to the balanced chemical equation:
[tex]\[
4 \text{ NH}_3(g) + 5 \text{ O}_2(g) \rightarrow 4 \text{ NO}(g) + 6 \text{ H}_2\text{O}(g)
\][/tex]
we follow these steps:
1. Calculate the number of moles of each reactant:
- Molar mass of NH₃ (ammonia) is 17.0 g/mol.
- Molar mass of O₂ (oxygen) is 32.0 g/mol.
[tex]\[
\text{Moles of NH}_3 = \frac{\text{Mass of NH}_3}{\text{Molar mass of NH}_3} = \frac{4.0 \, \text{g}}{17.0 \, \text{g/mol}} \approx 0.235 \, \text{moles}
\][/tex]
[tex]\[
\text{Moles of O}_2 = \frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{8.0 \, \text{g}}{32.0 \, \text{g/mol}} = 0.25 \, \text{moles}
\][/tex]
2. Determine the theoretical yield of NO from each reactant:
- According to the balanced equation, 4 moles of NH₃ produce 4 moles of NO.
- According to the balanced equation, 5 moles of O₂ produce 4 moles of NO.
[tex]\[
\text{Moles of NO produced by NH}_3 = \frac{\text{Moles of NH}_3}{\text{Coefficient of NH}_3} \times \text{Coefficient of NO} = \frac{0.235 \, \text{moles}}{4} \times 4 = 0.235 \, \text{moles}
\][/tex]
[tex]\[
\text{Moles of NO produced by O}_2 = \frac{\text{Moles of O}_2}{\text{Coefficient of O}_2} \times \text{Coefficient of NO} = \frac{0.25 \, \text{moles}}{5} \times 4 = 0.20 \, \text{moles}
\][/tex]
3. Identify the limiting reactant:
- The limiting reactant is the one that produces the least amount of product (NO).
- The moles of NO produced by NH₃ is 0.235 moles.
- The moles of NO produced by O₂ is 0.20 moles.
Since O₂ produces fewer moles of NO (0.20 moles), O₂ is the limiting reactant.
Thus, the correct answer is:
[tex]\[ \text{O}_2 \text{ because it produces only 0.20 mol of NO.} \][/tex]
