how many grams of water are needed to dissolve 27.8g of NH4NO3 in order to prepare a .452 m solution?
553883
27.8 g of ammonium nitrate = 0.348 moles (MW = 80)\n\nAssuming you mean molality and not molarity:\n\nkg solvent = 0.348 moles/0.452 m = 0.770 kg or 770 g
245442
Wait...not mixed with H2O? As in, molten liquid ammonia and molten liquid sodium sulfate? In that case no reaction. But with water, then:\n\n2NH3 + Na2SO4 + 2H2O --> (NH4)2SO4 + 2NaOH
756485
The water isn't involved in the reaction, so take it out.  Then, just put a 2 in front of the Nh2(SO3)H and a 2 in front of HNO3, and you're finished.\n\nNow, go take a nap or go to sleep...
921558
It will be clearer if you write the reaction as the sum of two reactions.\n\nWhen HN3 is dissolved in water, it reacts with the water to form ammonium ion (NH4+) and hydroxide ion (OH-).  Omitting the reaction NH3(g) = NH3(aq), we have:\n\nNH3 (aq) + H2O (L) = NH4+(aq) + OH-(aq)\n\nThe equilibrium constant for this reaction is:\n\nKb = [NH4+]*[OH-]/[NH3]\n\nThe hydroxide produced in this reaction reacts with the manganese ion to form insoluble manganese hydroxide (Ksp ~ 2*10^-13)\n\nMn++(aq) + 2OH-(aq) = Mn(OH)2(s)\n\nThe equilibrium constant for this reaction is:\n\nK = 1/Ksp = 1/([Mn++]*[OH-]^2) \n\nThe reaction you wrote is the sum of 2 times the first reaction plus the second reacton.  Note that as written, the first reaction produces OH- ions while the second one consumes them.  \n\nIf we add an acid like HNO3 to the solution, we are adding H+ ions.  Adding H+ ions is really like subtracting OH- ions because the concentrations of H+ and OH- are inversely proportional to one another through the water dissociation reaction (H2O = H+ + OH-, [H+]*[OH-] = Kw = 10^-14).\n\nWe can therefore write the two equilbrium constants above in terms of the concentration of H+ instead of OH-:\n\nKb = [NH4+]*[OH-]/[NH3] = Kw*[NH4+]/([NH3]*[H+])\n\nK = 1/([Mn++]*[OH-]^2) = ([H+]^2)/(Kw^2 * [Mn++])\n\nThis means that if we add acid (i.e., HNO3) to the solution, we will drive the first reaction in the forward direction; the concentration NH4+ will increase and the concentration of NH3(aq) will decrease.  Adding H+ will drive the second reaction in the reverse direction; Mn(OH)2(s) will dissolve, and the concentration of Mn++ will increase.
914609
9,50 g HCl*(1mol HCl/36,45g HCl)= 0,261 moles HCl\nM= 0,261 moles/0,0292L\n= 8,93 M
747027
It'd be basic speaking qualitatively. NH3 is a base so any amount in water would make the solution basic.
190428
22.4 l=>1 mole N2O\n0.1 l=> 0.0045 mole N20\nsince 1 mole NH4NO3 gives 1 mole N20,\n0.0045 mole of NH4NO3 is required.\nsince 1 mole NH4N03=>80 grams,\nmass of NH4N03 required is 80*0.0045=0.3571 grams\n\nsince 4 moles of Fe reacts with 3 moles of 02\nand 52 g of Fe=1 mole Fe,\n0.75 mole of 02 required.\nsince 1 mole 02=>22.4 l,\nvolume of 02 required is 16.8 litres\n\nhope that helps
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
924787
First of all\n3 moles of NO2 give you 2 moles of HNO3, while 1 mol of water gives you the same amount of product. We must find out which reactant gives you the least product, and that will be your maximum amount of product.\n36,0g H2O*(1mol H2O/18g H2O)*(2moles HNO3/1mol H2O)*(63g HNO3/1mol HNO3)= 252 g of HNO3\n\nNow with NO2\n92,0g NO2*(1mol NO2/46g NO2)*(2 moles HNO3/ 3moles NO2)*(63g HNO3/1mol HNO3)= 84,0g HNO3\nTherefore, the biggest amount of HNO3 you can expect to get is 84,0g.
960538
90.12g of H2O are produced. you have to first convert to molar mass, then use the mole ratio of the balanced equation for NH3 and H20, then use molar mass again.
1145574
0.150 M NaOH\n\n0.150mol / L  X  40 g / mol  X  1 L / 1000 mL  X  200 mL  =  1.2 grams of solute (NaOH)
294143
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
139469
16.45 mL HNO3 solution * [1 liter/1000 mL] * [0.302 moles HNO3/1 liter HNO3 solution] * [1 mol NaOH/1 mol HNO3] * [1 liter NaOH solution/0.250 mol NaOH] * [1000 mL/1 liter]\n\n=67.0 mL NaOH solution\n\nEDIT: [Sorry, the answer above was wrong; I don't know how I got that. The answer is 19.87 mL]\nEND EDIT\n\nBasically, you just use a whole bunch of ratios relating all the units and measurements together.\nIf you write all that out, all the units should cancel except the "mL NaOH solution"\nMight want to check my answer though...\n\n\nNext question:\n[16.45 ml HNO3 solution] * [1 liter/1000 ml] * [0.302 mol HNO3/1 liter HNO3 solution] * [1000 millimoles solution/1 mol solution]\n\n=4.97 millimoles HNO3 solution
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
955015
45.6 grams x 1 mol/323.2 g = 0.141 moles \n\n\nFormula weights \n\nPb= 207.2 x 1 = 207.2\nCr= 52 x 1 =         52.0\nO = 16 x4  =         64.0\n\nPbCrO4 =          323.2
935686
1. Look at the periodic table and figure out the formula mass of your compounds.  .005*formula mass of  Pb(NO3)2 and ditto for the 2nd one...\n\n2. 1/2 of 454g...
977807
M = moles/L\nso, 0.220M = 0.220 moles per liter and there are 2 liters so there are...\n0.440 moles of Mg(NO3)2\n\nBUT... notice that there are 2 ions of NO3 for each molecule, so double it again and get 0.880 moles of NO3