Question

consider the formation of nitrogen dioxide from nitric oxide and oxygen:
2 no(g) + o₂(g) → 2 no₂(g)
if 5.5 l of o₂ is combined with excess no at stp, what is the volume, in l, of no₂ produced? report your answer to one place after the decimal point.
note: r = 8.314 (\frac{kpa cdot l}{mol cdot k}), stp = 1 bar pressure, 273.15 k

Explanation:

Step1: Analyze the reaction stoichiometry

The balanced reaction is \( \ce{2 NO(g) + O2(g) -> 2 NO2(g)} \). From the stoichiometry, 1 mole of \( \ce{O2} \) produces 2 moles of \( \ce{NO2} \). At STP (same temperature and pressure), the volume ratio is equal to the mole ratio (Avogadro's law: \( V \propto n \) at constant \( T, P \)). So, the volume of \( \ce{NO2} \) produced is related to the volume of \( \ce{O2} \) by the ratio \( \frac{V_{\ce{NO2}}}{V_{\ce{O2}}} = \frac{2}{1} \).

Step2: Calculate the volume of \( \ce{NO2} \)

Given \( V_{\ce{O2}} = 5.5 \, \text{L} \). Using the volume ratio:
\( V_{\ce{NO2}} = 2 \times V_{\ce{O2}} \)
\( V_{\ce{NO2}} = 2 \times 5.5 \, \text{L} = 11.0 \, \text{L} \)

Answer:

11.0