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
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?
2 answers
good calculations but more quest3
5 answers