Dimethylamine burns in oxygen according to the following equation:
4 C2H7N + 15 O2 → 8 CO2 + 14 H2O + 2 N2
(a) How many liters of O2 at 35 °C and 0.840 atm will be needed to burn 7.70 L of C2H7N at 35 °C
and 0.840 atm?
(b) How many liters of CO2 at 35 °C and 0.840 atm will be produced?
Report your answers to parts (a) and (b) to 3 significant figures
To solve this problem, we will use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature (in Kelvin)
We'll start by converting the given conditions to Kelvin.
(a) To find the number of liters of O2 needed, we need to calculate the number of moles of C2H7N first.
Using the ideal gas law equation, we can rearrange it to solve for n:
n = PV / RT
Given:
P = 0.840 atm
V = 7.70 L
R = 0.0821 L·atm/(mol·K)
T = 35 °C = 35 + 273.15 = 308.15 K
Plugging the values into the equation:
n(C2H7N) = (0.840 atm * 7.70 L) / (0.0821 L·atm/(mol·K) * 308.15 K)
Calculate n(C2H7N):
n(C2H7N) ≈ 0.280 mol
Now, looking at the balanced equation, we can see that 4 moles of C2H7N react with 15 moles of O2. Therefore, the molar ratio is 15:4.
Using this ratio, we can calculate the number of moles of O2 required:
n(O2) = (4 mol of O2 / 15 mol of C2H7N) * 0.280 mol of C2H7N
Calculate n(O2):
n(O2) ≈ 0.0747 mol
Now, we can use the ideal gas law equation to find the volume of O2 in liters:
V = nRT / P
Given:
P = 0.840 atm
n = 0.0747 mol
R = 0.0821 L·atm/(mol·K)
T = 35 °C = 35 + 273.15 = 308.15 K
Plugging the values into the equation:
V = (0.0747 mol * 0.0821 L·atm/(mol·K) * 308.15 K) / 0.840 atm
Calculate V:
V ≈ 2.44 L
Therefore, approximately 2.44 liters of O2 will be needed to burn 7.70 L of C2H7N.
(b) To determine the number of liters of CO2 produced, we'll use the same approach, but this time using the molar ratio between CO2 and C2H7N.
According to the balanced equation, 8 moles of CO2 are produced for every 4 moles of C2H7N reacted.
Using this ratio, we can calculate the number of moles of CO2 produced:
n(CO2) = (8 mol of CO2 / 4 mol of C2H7N) * 0.280 mol of C2H7N
Calculate n(CO2):
n(CO2) ≈ 0.560 mol
Now, we can use the ideal gas law equation to find the volume of CO2 in liters:
V = nRT / P
Given:
P = 0.840 atm
n = 0.560 mol
R = 0.0821 L·atm/(mol·K)
T = 35 °C = 35 + 273.15 = 308.15 K
Plugging the values into the equation:
V = (0.560 mol * 0.0821 L·atm/(mol·K) * 308.15 K) / 0.840 atm
Calculate V:
V ≈ 12.9 L
Therefore, approximately 12.9 liters of CO2 will be produced.
To solve this problem, we need to use 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, and T is the temperature in Kelvin. We'll also need to use the stoichiometry of the chemical equation to convert between moles of reactants and products.
(a) To find the number of moles of O2 required, we first need to convert the volume of C2H7N to moles. We can use the ideal gas law equation to calculate the number of moles of C2H7N:
PV = nRT
n = PV / RT
Substituting the values given:
n = (0.840 atm * 7.70 L) / (0.0821 L·atm/(mol·K) * 308 K)
n ≈ 3.17 mol
According to the balanced equation, we need 15 moles of O2 for every 4 moles of C2H7N. Therefore, the number of moles of O2 required is:
n(O2) = (15/4) * n(C2H7N)
n(O2) ≈ (15/4) * 3.17
n(O2) ≈ 11.92 mol
Now, let's find the volume of O2 in liters using the ideal gas law equation:
V(O2) = n(O2) * (RT / P)
V(O2) = 11.92 mol * (0.0821 L·atm/(mol·K) * 308 K / 0.840 atm)
V(O2) ≈ 349 L
Therefore, approximately 349 liters of O2 are needed to burn 7.70 liters of C2H7N.
(b) To find the volume of CO2 produced, we use the same procedure as in part (a) but with the stoichiometric ratio for CO2:
According to the balanced equation, we produce 8 moles of CO2 for every 4 moles of C2H7N. Therefore, the number of moles of CO2 produced is:
n(CO2) = (8/4) * n(C2H7N)
n(CO2) = 2 * 3.17
n(CO2) ≈ 6.34 mol
Now, let's find the volume of CO2 in liters using the ideal gas law equation:
V(CO2) = n(CO2) * (RT / P)
V(CO2) = 6.34 mol * (0.0821 L·atm/(mol·K) * 308 K / 0.840 atm)
V(CO2) ≈ 231 L
Therefore, approximately 231 liters of CO2 will be produced.
To summarize:
(a) Approximately 349 liters of O2 will be needed.
(b) Approximately 231 liters of CO2 will be produced.
When working stoichiometric problems in an all gas system, one may use a shortcut in which volume may be used as mols directly. Therefore, all you need to do is to convert L of one material to L of the other; it is not necessary to go through mols.
For example:
7.70L BNH x (15 L O2/4 L BNH) = ? L O2 needed.
and
7.70 L BNH x (8 mols CO2/4 mols BNH) = ? L CO2 formed.
Note: It was convenient that the CONDITIONS (0.840 atm and 35 C) did NOT change. Had they changed we would have been forced to use PV = nRT to obtain mols.