Question
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?
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?
Answers
bobpursley
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
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
jean
good calculations but more quest3