Consider the reaction for the production of NO2 from NO:
2 NO(g) + O2(g) = 2 NO2(g)
a)If 84.8L of O2(g), measured at 35 degrees Celsius and 632mm Hg, is allowed to react with 158.2g of NO, find the limiting reagent.
b) If 97.3L of NO2 forms, measured at 35 degrees Celsius and 632mm Hg, what is the percent yield?
a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Step 2. Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
% yield = (AY/TY)*100 = ?
b) 60%
a) To determine the limiting reagent, we need to compare the amount of NO and O2 and see which one will run out first.
1. Convert the volume of O2 from liters to moles using the ideal gas law equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Since the temperature is given in degrees Celsius, we need to convert it to Kelvin by adding 273.15.
T = 35 + 273.15 = 308.15K
Now, plug in the values:
(P O2) (V O2) = (n O2) (R) (T)
(632 mmHg) (84.8 L) = (n O2) (0.0821 L·atm/mol·K) (308.15 K)
Convert mmHg to atm by dividing by 760:
(0.832 atm) (84.8 L) = (n O2) (0.0821 L·atm/mol·K) (308.15 K)
Simplify:
(70.2976) = (n O2) (25.30778)
n O2 = 2.776 mol O2
2. Convert the mass of NO to moles:
m NO = 158.2 g
Molar mass of NO = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
n NO = (158.2 g) / (30.01 g/mol) = 5.271 mol NO
3. Use the balanced equation to compare the ratio of NO to O2:
2 NO : 1 O2
5.271 mol NO : (2.776 mol O2)/2 = 1.388 mol O2
Since the ratio is less than 1, the limiting reagent is O2.
b) To calculate percent yield, we need the actual yield and the theoretical yield. The theoretical yield can be calculated using the limiting reagent:
1. Convert the volume of NO2 from liters to moles using the ideal gas law equation, same as in part a:
(P NO2) (V NO2) = (n NO2) (R) (T)
(632 mmHg) (97.3 L) = (n NO2) (0.0821 L·atm/mol·K) (308.15 K)
Convert mmHg to atm by dividing by 760:
(0.832 atm) (97.3 L) = (n NO2) (0.0821 L·atm/mol·K) (308.15 K)
Simplify:
(80.7776) = (n NO2) (25.30778)
n NO2 = 3.195 mol NO2
2. Use the stoichiometry of the balanced equation to calculate the theoretical yield of NO2:
2 mol NO : 2 mol NO2
5.271 mol NO : (3.195 mol NO2)/2 = 2.081 mol NO2
3. Calculate the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) x 100%
Actual Yield = 97.3 L (given)
Theoretical Yield = 2.081 mol NO2 x (44.01 g/mol) = 91.44 g NO2
Percent Yield = (97.3 g / 91.44 g) x 100%
To find the limiting reagent in the given reaction, we need to compare the amounts of each reactant provided and determine which one will be completely consumed first.
a) To find the limiting reagent, we will use the concept of stoichiometry and molar ratios. First, we need to convert the given quantities of O2 and NO into moles.
1. Convert the volume of O2 to moles using the ideal gas law equation: PV = nRT, where
P = pressure (632 mm Hg)
V = volume (84.8 L)
n = moles
R = gas constant (0.0821 L.atm/K.mol)
T = temperature in Kelvin (35 degrees Celsius + 273.15)
Using PV = nRT, we have:
n(O2) = (P * V) / (R * T)
n(O2) = (632 mm Hg * 84.8 L) / (0.0821 L.atm/K.mol * (35 + 273.15))
n(O2) = 22.31 mol
2. Convert the mass of NO to moles using its molar mass. The molar mass of NO is 30.01 g/mol.
n(NO) = mass / molar mass
n(NO) = 158.2 g / 30.01 g/mol
n(NO) = 5.27 mol
Now that we have the number of moles for each reactant, we can compare their stoichiometric coefficients to determine which reactant is limiting. The stoichiometric coefficient ratio is 2:1 for NO/O2.
Comparing the moles, we find that we have excess O2 (22.31 mol) and limited NO (5.27 mol). This means NO is the limiting reagent.
b) To find the percent yield, we need to compare the actual amount of NO2 formed to the theoretical amount calculated from the limiting reagent.
1. Calculate the theoretical yield of NO2 using the stoichiometric ratio from the balanced equation. The stoichiometric ratio is 2:2, so the number of moles of NO2 formed will be equal to the number of moles of NO used.
Theoretical yield of NO2 = moles of NO2 formed = moles of NO used = n(NO)
Theoretical yield of NO2 = 5.27 mol
2. Convert the volume of NO2 produced to moles using the ideal gas law equation. The conditions are the same as given in part (a) (35 degrees Celsius and 632 mm Hg).
n(NO2) = (P * V) / (R * T)
n(NO2) = (632 mm Hg * 97.3 L) / (0.0821 L.atm/K.mol * (35 + 273.15))
n(NO2) = 32.76 mol
3. Calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (32.76 mol / 5.27 mol) * 100
Percent Yield = 622.96%
Therefore, the percent yield of NO2 is 622.96%.