To solve this problem, we need to determine the theoretical yield of [tex]\( NO_2 \)[/tex] produced when 50.0 L of [tex]\( NO \)[/tex] and 50.0 L of [tex]\( O_2 \)[/tex] are combined.
Step-by-Step Solution:
1. Balanced Chemical Equation:
The balanced chemical equation is given as:
[tex]\[
2 NO (g) + O_2 (g) \rightarrow 2 NO_2 (g)
\][/tex]
2. Determine the Stoichiometric Ratios:
- According to the balanced equation, 2 volumes of [tex]\( NO \)[/tex] react with 1 volume of [tex]\( O_2 \)[/tex] to produce 2 volumes of [tex]\( NO_2 \)[/tex].
3. Identify the Limiting Reactant:
- To identify the limiting reactant, we need to compare the ratio of the given volumes to the stoichiometric coefficients from the balanced equation.
- The required volume ratio of [tex]\( NO \)[/tex] to [tex]\( O_2 \)[/tex] is 2:1.
4. Calculate the Volume Ratios:
- For [tex]\( NO \)[/tex]:
[tex]\[
\text{Ratio}_{NO} = \frac{\text{Volume of } NO}{\text{Stoichiometric coefficient of } NO} = \frac{50.0 \text{ L}}{2} = 25.0
\][/tex]
- For [tex]\( O_2 \)[/tex]:
[tex]\[
\text{Ratio}_{O_2} = \frac{\text{Volume of } O_2}{\text{Stoichiometric coefficient of } O_2} = \frac{50.0 \text{ L}}{1} = 50.0
\][/tex]
5. Determine the Limiting Reactant:
- The smaller ratio indicates the limiting reactant. Since [tex]\( 25.0 < 50.0 \)[/tex], [tex]\( NO \)[/tex] is the limiting reactant.
6. Calculate the Theoretical Yield of [tex]\( NO_2 \)[/tex]:
- According to the stoichiometry of the balanced equation, 2 volumes of [tex]\( NO \)[/tex] produce 2 volumes of [tex]\( NO_2 \)[/tex].
- Since we have 50.0 L of [tex]\( NO \)[/tex] as the limiting reactant, the volume of [tex]\( NO_2 \)[/tex] produced will be equal to the volume of [tex]\( NO \)[/tex] used, which is 50.0 L.
Therefore, the theoretical yield for the volume of [tex]\( NO_2 \)[/tex] formed at STP is:
[tex]\[
50.0 \text{ L NO}_2
\][/tex]
