Nitroglycerin, C3H5(NO3)3, a solid explosive compound, decomposes to carbon dioxide, water vapor, nitrogen, and oxygen. (a) Calculate the total volume of gases when collected at 1.2 atm and 25 degrees Celsius from 2.6*10^2 g of nitroglycerin. (b) What are the partial pressures of the gases under these conditions?
I will just put the work I did for (a) right below, in case anything I did in between is relevant to use in (b). Otherwise you can just skip all the way to long arrow, where I state my confusion for (b).
So for (a), I did:
Balanced equation:
2C3H5(NO3)3 -> 6CO2 + 5H20 + 3N2 + O2
The molar mass of C3H5(NO3)3 = 227.11 g/mol
Conversion of g to mol of C3H5(NO3)3 in the decomposition reaction:
2.6*10^2g(1mol/227.11g) = 1.144819691 mol = 1.14 mol
Based on the balanced equation, the ratios are:
2mol C3H5(NO3)3 : 6mol CO2 : 5mol H20 : 3mol N2 : 1mol O2
So for mol of products produced, using stoichiometric ratios:
2*mol of C2H5(NO3)3 = 3(1.14mol) = 3.434459073mol = 3.43 mol CO2
(5/2)(1.14mol) = 2.862049228mol = 2.86 mol H20
(3/2)(1.14mol) = 1.717229537mol = 1.72 mol N2
(1/2)(1.14mol) = 0.5724098mol = 0.572 mol O2
Then volume calculations for each gas produced:
T = 307degreesC + 273.15 = 298.15K = 298K
PV = nRT -> V = nRT/P
CO2: V = (3.43mol)(0.0821Latm/molK)(298K)/(1.2atm) = 70.05757013L = 70.1L
H20: V = (2.86mol)(0.0821Latm/molK)(298K)/(1.2atm) = 58.38130845L = 58.4L
N2: V = (1.72mol)(0.0821Latm/molK)(298K)/(1.2atm) = 35.02878507L = 35.0L
O2: V = (0.572mol)(0.0821Latm/molK)(298K)/(1.2atm) = 11.6762617L = 11.7L
Total volume of gases produced = 175.1439254L = 175L
------------------->Now for (b), I don't know what to do... if the pressures for all of the gases are 1.2 atm?
partial pressure for each is related directly to the mole fraction
since you have the volumes totaled, the mole fraction is the same as
take O2 for instance: molefraction=11.7/175
partial pressure O2= (molefractio)1.2 atm
good calculations but more quest3
For part (b), you need to calculate the partial pressures of each gas under the given conditions (1.2 atm and 25 degrees Celsius).
To do this, you will need to use the ideal gas law equation, which is:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
First, convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Next, calculate the partial pressures for each gas using the number of moles and the volume calculated in part (a).
For CO2:
n = 3.43 mol
V = 70.1 L
P = (n * R * T) / V = (3.43 mol * 0.0821 L·atm/mol·K * 298.15 K) / 70.1 L = 1.45 atm
For H2O:
n = 2.86 mol
V = 58.4 L
P = (n * R * T) / V = (2.86 mol * 0.0821 L·atm/mol·K * 298.15 K) / 58.4 L = 1.48 atm
For N2:
n = 1.72 mol
V = 35.0 L
P = (n * R * T) / V = (1.72 mol * 0.0821 L·atm/mol·K * 298.15 K) / 35.0 L = 2.06 atm
For O2:
n = 0.572 mol
V = 11.7 L
P = (n * R * T) / V = (0.572 mol * 0.0821 L·atm/mol·K * 298.15 K) / 11.7 L = 4.19 atm
Therefore, the partial pressures for CO2, H2O, N2, and O2 under these conditions are 1.45 atm, 1.48 atm, 2.06 atm, and 4.19 atm, respectively.
To calculate the partial pressures of the gases in part (b), you need to use Dalton's Law of Partial Pressures. According to Dalton's Law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases.
In this case, you have calculated the total volume of the gases (175 L) in part (a). Now, you need to find the mole fractions of each gas to determine their partial pressures.
To find the mole fraction of a gas, you divide the moles of that gas by the total moles of all the gases.
Mole fraction (X) = Moles of gas / Total moles of all gases
The total moles of all gases can be calculated by adding up the moles of CO2, H2O, N2, and O2. In this case, the total moles of all gases is 3.43 mol + 2.86 mol + 1.72 mol + 0.572 mol = 8.584 mol.
Now, let's calculate the mole fractions of each gas:
Mole fraction of CO2 = 3.43 mol / 8.584 mol = 0.399
Mole fraction of H2O = 2.86 mol / 8.584 mol = 0.333
Mole fraction of N2 = 1.72 mol / 8.584 mol = 0.200
Mole fraction of O2 = 0.572 mol / 8.584 mol = 0.067
The partial pressure of each gas can be found by multiplying the mole fraction by the total pressure. In this case, the total pressure is given as 1.2 atm.
Partial pressure of CO2 = 0.399 * 1.2 atm = 0.479 atm
Partial pressure of H2O = 0.333 * 1.2 atm = 0.400 atm
Partial pressure of N2 = 0.200 * 1.2 atm = 0.240 atm
Partial pressure of O2 = 0.067 * 1.2 atm = 0.080 atm
Therefore, the partial pressures of CO2, H2O, N2, and O2 are 0.479 atm, 0.400 atm, 0.240 atm, and 0.080 atm, respectively.