ow many grams of   H2O   are produced from the complete reaction of  68.12 g   of   NH3 ?
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.
959309
#3:   113.6g ammonia is the maxium that can be produced, i think.\n\nBecause:   when you set up the problem you do:   5 moles N2 x 2 moles of NH3 (coming from the balanced equation) x 17.04g NH3 (formula mass) equaling.....170.4g NH3\n\nthen....\n\nyou have to do it again with the other number(10 moles of H2)\n\n10 moles H2 x 2 moles NH3 / 3 moles H2 x 17.04 equaling....113.6.   \n\nyou pick the lowest number becuase that is the maxium you can have.   so your answer is 113.6g NH3.\n\n\n\n#4:   15.33g NO2 is the maxium that can be produced \n\nbecause: 20g of O2 to grams of NO2.  To do that, you do...20g O2 / 32g O2 (Formula mass of O2) x 46.0g NO2 (formula mass of NO2)\nwhich equals......28.76\n\nthen do:\n\n10gNO / 30.0g NO (formula mass of NO) x 46.01(formula mass of NO2) and that equals....15.33\n\nso your answer is the lowest number which is 15.33.\n\n\ni hope that helps and i hope i'm right...
573821
I think this question is incomplete because the Ka of HF hasn't been provided.  The amount of H+ that is formed from HF can only be calculated from its Ka value.
773743
The key word in here is the "excess nitrogen". Having said that means you will provide ENOUGH or whatever amount it takes nitrogen for the reaction. i.e., H2 is your limiting reagent.\n\nN2  + 3H2  ---->  2NH3\n\nWith 5.74 g H2,\n# of moles of H2 = 5.74 / 2 = 2.87 moles\n\nAccording to the balanced equation, every 3 moles of H2 will produce 2 mole of NH3 (provided that you supply enough N2). So,\n# of moles of NH3 produced (theoretically) = (2.87/3)*2 = 1.913 moles\n\nmass of NH3 produced (theoretically) = 1.913*(14 + 3) = 32.527 g\n\nActual yield of NH3 from the process is 21.3 g\n\n% yield = (actual yield / theoretical yield) * 100\n= (21.3 / 32.527) * 100 = 65.48%
49738
C*6+H*8+O*6=\n12.01*6+1.008*8+16.00*6=176.12 g/mol
664234
NH4ClO4 --> NH4^+ + ClO4^-\n\nNow in a water solution,\nClO4^- + H2O --> HClO4 + OH- --> This doesn't happen because HClO4 is a strong acid.  Therefore the only contribution to the pH is from the NH4^+\n\nNH4^+ <=> NH3 + H^+\nYou start with [NH4^+] = 0.40 M, and [NH3]=[H^+] = 0 M; after undergoing equilibrium [NH4^+] = 0.40 - x, while [NH3]=[H^+] = x; Ka = [NH3][H^+]/[NH4^+] = (x)(x)/(0.40-x) = Kw/Kb = (1 x 10^-14)/(1.8 x 10^5)\nAssume x << 0.40, so x^2/0.40 = 5.56 x 10^-10\nx = 1.49 x 10^-5 M = [H^+]\npH = -log(1.40 x 10^-5) = 4.827
860532
nitrogen & hydrogen react to give ammonia according to the reaction \nN2 + 3H2 -> 2NH3\naccording to the above reaction 1 & 3 moles of N2 & H2 respectively react to give 2 moles of NH3.\nso, 6 moles ammonia can be produced by reacting 3 moles N2 & 9 moles H2.
813561
The molecular mass of NaOH is 40.00 gm/mol,  so 8 gm of NaOH contains 8 gm/40.00 gm.mol = 0.20 moles of NaOH.\n\nThe molecular mass of HCl is 36.46 gm/mol, so 8 gm of HCl contains 8gm/36.46 gm/mol = 0.22 moles of HCl.\n\nIf we react these amounts of NaOH and HCl, the NaOH will be the limiting reactant, and we will produce 0.20 moles each of NaCl and H2O.  This reaction will liberate \n\n0.20 mol * 56 kJ/mol = 11.2 kJ of energy as heat\n\nThe heat capacity of the final solution is given by the mass of the final solution (which, by conservation of mass, is equal to the mass of the initial solutions) multiplied by the specific heat capacity of the final solution.  With the assumptions given in the question, the heat capacity of the final solution is:\n\n(58 gm + 258 gm) * 4.184 J/(gm*degC) = 1322 J/degC.\n\n(Note that we don't need to use the density of the final solution.)\n\nThe temperature change of the solution is given by the energy released divided by the heat capacity of the system:\n\n(11.2*10^3 J)/(1322 J/degC) = 8.47 deg C\n\nIf the initial solutions had a temperature of 25 C, the final solution has a temperature of 25 + 8.47 C = 33.47 C.
750432
Amount of N2 which reacted = 21/ relative molecular mass of N2\n                                               = 21/ 28\n                                               = 0.75mol.\nMass of H = (0.75 x 6/2) x 2 = 4.5g of H2\nMass of NH3 = (0.75 x 2) x 17 =25.5g of NH3.\n\nThe value in the parentheses are your mol values of H and NH3 respectively.
753121
Ok, now this one I can do:\n\nMg3N2  ---> 3Mg (s) + N2 (g)
452634
N2 +3H2 ---> 2NH3\n\nso, H2 Mwt is 2g/mol and NH3 Mwt is 34g/mol\nas per above eqn, 6g of H2 will give u 34g of NH3\nSo, 9.988g of H2 will give 56.6g of NH3\n\nYou got it?
638948
The reaction NaOCl + NH3 = NH2Cl + NaOH is very easy even cold. The reaction NH2Cl + NH3 + NaOH = N2H4 + NaCl + H2O is not so easy as it requires heating to boiling or close to it.\n\nHere's the link: http://www.sciencemadness.org/talk/post.php?action=reply&fid=2&tid=1128&repquote=20454
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
937526
If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)\nFill in the atomic masses of H = 1.00797 and N = 14.0067:\n[3 x 2 x (1.00797)] +[ 2 x (14.0067)]  = 2 x (14.0067) x 3 x (1.00797) = 34.06122\nThis is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:\n2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.
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.
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
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)