NH3 + O2     →     NO   +   H2O how many grams of   NO are produced from the complete reaction of 64.00 g  o2
832542
Balance the Equation\n4NH3 + 5O2 --> 4NO + 6H2O\n\nConvert from g O2 to moles O2\n\n(64.00 g O2) * (1 mol O2)/(32.00 g) = 2 mol O2\n\nConvert from moles O2 to moles NO\n\n(2 mol O2) * (4 mol NO)/(5 mol O2) = 1.6 mol NO \n\nConvert from moles NO to grams NO\n\nbut first we need to figure out NO's molecular weight\n\n1 mol N * 14.01g/mol = 14.01g\n1 mol O * 16.00g/mol = 16.00g\n                                        30.01g/mol NO\n\nNow we can continue\n\n1.6 mol NO * 30.01g/mol = 48.02 g NO produced\n\nHope this helps!
956489
2 SO2 + O2 --> 2 SO3\n\n\nok, 64.00 g oxygen times the molar mass of 02( 32.00) to get mols of O2 then time the mole ratio of O2 to SO2 which is (2 mol SO2/2 Mol O2) and then u have the mols of SO2 times that by the molar mass of SO2 to get the mass of SO2\n\n64.00 x 32.00 x 2/2 x molar mass of SO2
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
793971
According to the equation:\n\n2NO + O2 --> 2NO2\n\n2 mol of NO reacts with 1 mol O2, so\n0.886 mol NO reacts with 0.886/2 = 0.443 mol O2\n\nWe have 0.503 mol O2 > 0.443 mol. So O2 is in excess and NO is the limiting reactant. It reacts fully and produces 0.886 mol NO2 (because 2 mol NO give 2 mol NO2)
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.
892172
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
55863
NH4(OH)\n\nIt is from combination of NH4+ , OH-
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.
954211
Your first conclusion is correct, but your second isn't.  Since you already have the mechanism given to you, you can see that the 2nd step is the slow step.  So, the rate law should be:\n\nrate = k (N2O2) (H2)\n\nSince N2O2 is an intermediate, it can't appear in the overall rate law, so you have to get rid of that.  From the first reaction, you see that N2O2 is formed from the collision of 2 NO.  So, the concentration of N2O2 is proportional to [NO]^2.  That gives you the overall rate law that you had, and the overall order of the reaction is 3rd order.\n\nYour second conclusion is incorrect because the slow step requires the collision of N2O2 and H2.  The slow step is just a second order reaction.  \n\nSo, you've got things pretty well in hand...Hope this helps a little...
326922
Another easy reaction:\n\nMg(s) + HNO3 ===========> Mg(NO3)2 + H2 (g)\n\nMagnesium reacts rapidly with nitric acid to make magnesium nitrate and hydrogen gas
605005
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
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...
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
233434
This is just a simple stoichiometry problem.  You have 200g of NH3, your first job is to calculate how many moles of NH3 this is.  From the periodic table, find the atomic weights of N and H to calculate the molar mass of NH3:  \n\nN=14 and H=1, so NH3 = 17.  \n\n200g NH3 X (1 mol NH3 / 17g NH3) = 11.76 mol NH3\n\nOkay, now that we know we're working with 11.76 moles of NH3, we can use the mole ratio from the given chemical equation to figure out how many moles of oxygen are needed.  Look at the coefficients in the equation -- we need 5 moles of O2 for every 4 moles of NH3:\n\n11.76 mol NH3 X (5 mol O2)/(4 mol NH3) = 14.7 mol O2\n\nNow we know we need 14.7 mol O2.  All we have to do now is convert that to grams, again using the atomic mass from the periodic table.  The atomic mass of O is 16, so O2=32  \n\n14.7 mol O2 X 32g/mol = 470.4 g O2\n\nSUMMARY:\nConvert grams ammonia to moles ammonia using atomic weights.\nUse the mole ratio to determine needed moles of O2.\nConvert moles O2 to grams O2 using atomic weights.
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
1006740
The equilibrium constant for this reaction is given by:\n\n[NO]^2/([N2]*[O2]) = 6.91*10^-4\n\nYou also know, from mass balance, that:\n\n[NO] + 2*[N2] = 2*0.75 mol/L \nand\n[NO] + 2*[O2] = 2*2.4 mol/L\n\n (i.e., the total concentration of nitrogen and oxygen atoms doesn't change in this system -- the number of nitrogen atoms at equilibrium after the gasses react must be the same as the number of atoms you started with)\n\nYou now have three equations in three unknowns ([NO], [O2], and [N2]).  These three unknowns are the concentrations of the gases at equilibrium.  \n\nTo solve these equations, we can rewrite the mass-balance equations as:\n\n[N2] = 0.75 - 0.5*[NO]\n[O2] = 2.4 - 0.5*[NO]\n\nsubstitute these into the equilbrium constant equation to eliminate [N2] and [O2]:\n\n[NO]^2/((0.75 - 0.5*[NO])*(2.4 - 0.5*[NO])) = 6.91*10^-4\n\n[NO]^2/{1.8-1.575*[NO]+0.25*[NO]^2} = 6.91*10^-4\n\n[NO]^2 = 0.0012438 -0.001088325*[NO] +0.00017275*[NO]^2\n\n0 = 0.0012438 - 0.001088325*[NO] - 0.99982725*[NO]^2\n\nUse the quadratic formula to solve for [NO].  In general, there will be two roots, but only one will be physically meaningful.\n\n[NO] = -0.03582 or 0.03473.  Obviously, only the positive root has a physical meaning, so at equilibrium [NO] = 0.03473 mol/L\n\nSubstitute this into the mass-balance equations to get the concentrations of [O2] and [N2]:\n\n[N2] = 0.75 - 0.5*0.03474 = 0.73263 mol/L\n[O2] = 2.4 -  0.5*0.03474 = 2.38263 mol/L\n\nYou can check these results by plugging them back into the equilibrium constant equation to make sure everything agrees:\n\n0.03473^2/[0.73263*2.38263] ?= 6.91*10^4\n\n6.91*10^-4 = 6.91*10^-4  Check!
937793
You need to use the Ideal Gas Law, PV=nRT, to find the number of moles of NH3 and O2. NH3 and O2 react in a 4:5 ratio, find out which is the limiting reagent and which is in excess. You'll find that O2 is the limiting reagent. Then for every 5 moles O2 reacted, 4 moles NO are produced. When you know the number of moles of NO produced use the MW of nitrogen monoxide to find the grams NO produced.\n\nfor STP, T=0 deg C\n               P=1 atm