Consider the reaction for the production of [tex] \text{NO}_2 [/tex] from [tex] \text{NO} [/tex]:

\[ 2 \text{NO(g)} + \text{O}_2\text{(g)} \rightarrow 2 \text{NO}_2\text{(g)} \]

(a) If 84.8 L of [tex] \text{O}_2\text{(g)} [/tex] is measured at 35°C and 632 mm Hg, and it is allowed to react with 158.2 g of [tex] \text{NO} [/tex], find the limiting reagent.

(b) If 97.3 L of [tex] \text{NO}_2 [/tex] forms, measured at 35°C and 632 mm Hg, what is the percent yield?

Additionally, draw a Lewis structure for each molecule and determine its molecular geometry. What kind of intermolecular forces are present in each substance?

- (a) [tex] \text{BCl}_3 [/tex] (remember that B is a frequent exception to the octet rule)

so now we use the number of moles of NO 5.27 moles. use again the PV=nRT
R=62.36367(L mmHg K−1 mol−1)
T=35C + 273 = 308K
P=632 mm Hg
V=97.3 L
n=632*97.3/62.3636367*308=3.20145770868
Look at the mole ratio of NO to NO2 its 2:2 they share the same number of moles.
Atomic weight of each element in NO2
N:14 O:16
Molar Mass:(16*2)+(14*1)=46 g
Actual number of moles=3.20
Theortical number of moles=5.20
Percentage Yield=(Actual Yield/Theortical Yield)*100