Mass of Produced Ammonia?
149153
This is a limiting reagent problem. You must calculate the moles of NH3 produces from the given amount of grams of N2 and H2.\n\n3.00g N2 x 1mol/28.02g x 2 mol NH3/1 mol N2 = .214 mol NH3\n\n.54g H2 x 1 mol/2.02g H2 x 2 mol NH3/3 mol H2 = .178 mol NH3\n\nFrom this calculation you now know that the H2 is the limiting reagent. The next part to calculate the grams of NH3 goes like this:\n\n.178 mol NH3 x 17.04g/1 mol = 3.03 g NH3
747027
It'd be basic speaking qualitatively. NH3 is a base so any amount in water would make the solution basic.
457480
(NH4)2CO3 is Ammonium Carbonate
699905
mass initial = Mi\nmass final = Mf\n\nMass growth = Mf - Mi\nPercent growth = Mf/Mi * 100
860173
The answers for your questions must be determined experimentally.  I can only give you some theory behind the answer below.\n\nCalcium chloride can react with ammonia gas to gaive a stable compound with the formula CaCl2*8NH3\n\nThis means one mole of Calcium chloride will adsorb 8 moles of Ammonia.  This means that about 111 grams of anhydrous (bone dry) calcium chloride will adsorb 8 (17) = 136 grams of Ammonia.\n\n136 g / 454 g/lb = 0.300 lb of Ammonia for about every 111 g / 454 g/lb = 0.25 pounds of dry Calcium chloride.\n\nI am sure that the electrical resistance would change but you would need to take experimental readings to see how much.\n\nA mixture of CaCl2 plus CaBr2 seem to be better for Ammonia storage, but may be more expensive.
852232
Pv = nRT\nn = m/M (M Molecular weight)\nPV = (m/M)RT\nP = (m/V)RT/M\nPM/RT = m/v = Density.\nPlug the numbers, watch the units and get the density.\n\nOh a numerical answer, yes, ...42.
436267
http://www.lloydminsterheavyoil.com/nitrogen.jpg\nhttp://www.tcm.phy.cam.ac.uk/castep/MD/img43.png
438289
Co(NH3)4 would be more likely
877279
First you need to determine your limiting reagent in the first step.\n\nMolecular mass of NH3 = 17 g/mol, O2 = 32 g/mol\nMol of NH3 = 5,700,000 g/17 g/mol\nMol of NH3 = 3.35E5 mol\n\nMol of O2 = 57,000,000 g/32 g/mol\nMol of O2 = 1.78E6 mol\n\nSo based on the first equation (1), NH3 is the limiting reagent.\nFor every 4 mol of NH3, one gets 4 mol of NO.  So the first reaction produced 3.35E5 mol of NO.\n\nThe second reaction produces 2 mol of NO2 for every 2 mol of NO used, so the second reaction produced 3.35E5 mol of NO2.\n\nThe third reaction produces 2 mol of HNO3 for every 3 mol of NO2 used, so this final reaction will produce 2/3 x 3.35E5 mol of HNO3 = 2.23E5 mol.\n\nThen use molecular mass of HNO3 to calculate how much of it (mass) that you made.\nHNO3 molecular mass = 63 g/mol\nmass = mol x molar mass\nmass = 2.23E5 x 63 g/mol\nmass = 1.41E7 g of HNO3\n\nOf course this assumes 100% yields in all the above noted reactions.
178523
Although I know I am not getting this $20.  199 kJ of energy.\n\nIn response to a later answer that claimed I was wrong he was right and wrong\n\n He wrote "first find the amount of mol of nitrogen\nmol = 1/14" But infact the amount was 1 kg and that divided by 14g/mol would be 71.43 moles and multiplying that times his enthalpy of vaporization which is 2.7928kJ/mol would give 199 kJ of energy. So I was right, but wrong with my significant figures because the asker said it was 1.0 kg which would gie me 2 sig figs and the answer would be 2.0*10^2 kJ.
1104610
Ammonia is shipped as a liquified gas under its own vapor pressure of 114 psig. The oldest commercial refrigerant known and still in use today. Most extensive use is in soil fertilization. This application is used in the form of salts, nitrates and urea.\n\nIt is the simplest stable compound of these elements and serves as a starting material for the production of many commercially important nitrogen compounds. Pure ammonia was first pre-pared by Joseph Priestley in 1774, and its exact composition was determined by Claude-Louis Berthollet in 1785.
439393
The nitrogen gas in air gets converted to 1) ammonia 2) nitrates (from nitrites)
159316
[Ca(NO3)2]=n/V  ---> n=[]*V= 0.390M * 0.044L=0.01716 mol\nn=m/M ---> m=n*M= 0.01716mol * 164 g/mol=2.81424 g
497226
2NH4V03 + H2S04 ---> (NH4)2S04 + 2V02 + H20
750464
My chemistry is pretty rusty, but let's see.\n\nI'll assume that you have an infinite amt of O2, and that you have 68.12 grams of NH3.\n\nThe equation isn't balanced, so you need to do that first (you have 3 hydrogens on one side, but two on the other).  Balance to get:\n\n2(NH3) + (2.5)(O2) --> 2NO + 3(H20)\n\n4(NH3) + 5(O2) --> 4NO + 6(H20)\n\nNow, given that you have 68.12 grams of NH3, you need to convert this into moles of NH3, as Stoichiometry deals with molar ratios.  To get the conversion ratio, look at the periodic table.  Under each element, there should be an atomic mass.  For oxygen is 16.something, Hydrogen 1.008 ( i think ).  Find it for N and H.  (this is where I get hazy on the math).  I believe: take 3 * atomic mass of hydrogen + atomic mass of nitrogen, let's call it M  (it should be in grams/mole)  So to get moles, multiply 68.12 * (1/M).  \n\nNow, our equation above shows that for every 4 moles of NH3, we get 6 of H20.  # moles NH3 * (6/4) = # moles of H20.  Now, do the reverse operation of H20 to get how many grams is produced.  H20 has an atomic mass of something like 18.016.  So, # moles H20 * 18.016 = # grams H20\n\n\nI hope that's somewhat correct, and if not, leads you in the right direction at least
623526
Nitrogen (approx 78%)\nOxygen (approx 21%)
860730
Multiply moles by the molecular weight (grams per mol) to get mass (grams).