Question

If someone can walk me through these
116. Consider the reaction for the production of NO2 from NO:
2 NO(g) + O2(g)¡2 NO2(g)
(a) If 84.8 L of O2(g) measured at 35 °C and 632 mm Hg, is allowed to react with 158.2 g of NO, find the limiting reagent.
(b) If 97.3 L of NO2 forms, measured at 35 °C and 632 mm Hg, what is the percent yield?
• Exercise 92 in Ch. 12 (Page 444)
92. Draw a Lewis structure for each molecule and determine its molecular geometry. What kind of intermolecular forces are present in each substance?
(a) BCl3 (remember that B is a frequent exception t

Answer 1

Use ideal gas equation
PV=nRT
P- Pressure V-Volume n-Number of moles R-Ideal gas constant T-Temperature (measured in kelvin). we change 35 Degree Celsius to Kelvin by adding to 273. T = 35C + 273 = 308K R=62.36367(L mmHg K−1   mol−1)
http://en.wikipedia.org/wiki/Gas_constant
n=PV/RT, n = (632mmHg)(84.8L)/(62.36)(308K) = 2.79 moles O2
Number of Moles=Mass(g)/Molar Mass(g/mol)
Number of Moles of NO=158.2/30=5.27 moles of NO.
Balance Chemical equation
\[2NO(g) + O2(g)\rightarrow2 NO2(g) \]
Look at the mole ratio and if you find the calculating number of moles more than the actual moles this will be the limiting reagent.

Answer 2

You can find the limiting reagent directly by dividing the moles of each reactant by their corresponding stoichiometric coefficient and comparing them.

\(\sf \dfrac{moles~ofO_2}{coefficient}=\dfrac{2.79 ~moles }{1}\)

\(\sf \dfrac{moles~of~NO}{coefficient}=\dfrac{5.27~moles}{2}\)

Whichever is less after the division is the limiting reagent.

Answer 3

ok so NO would be the reagent. Right? O2=2.76 NO=2.635

Answer 4

yes NO is the limiting reagent. 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

Answer 5

Thank y'all so much for your help

Answer 6

your most welcome :)