Consider the following reaction for the production of NO2.
2 NO(g) + O₂(g) → 2 NO₂(g)
What mass of O₂ is required to produce 5.00 L NO₂ at 305 K and 752 mm Hg? (R = 0.08206 L⋅atm/mol⋅K)

Question

Quizplus · Accepted Answer

First, use the ideal gas law to find moles of NO2: PV = nRT. Convert pressure to atm (752 mm Hg = 0.989 atm). n = (0.989 atm * 5.00 L) / (0.08206 L⋅atm/mol⋅K * 305 K) = 0.197 mol NO2. From the balanced equation, 1 mol O2 produces 2 mol NO2, so moles of O2 needed = 0.197 mol / 2 = 0.0985 mol. Mass of O2 = 0.0985 mol * 32.00 g/mol = 3.16 g.

Consider the Following Reaction for the Production of NO2