what volume of 0.75 mol.L NH4 NO3 would be required to prepare 450 mL of 0.30 mol/L NH4 NO3?
88407
M1V1 = M2V2.  You know three of the four variables.
438289
Co(NH3)4 would be more likely
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%
829104
Well, it's 0.205 moles/liter and you're making 1/4 liter, so you need 0.205/4 or 0.05125 moles.  The molar mass of NaOH is equal to the sum of the molar masses of Na, O, and H, or 22.990 + 15.999 + 1.007 = 39.996 g/mole.  The mass on NaOH needed is equal to 0.05125mole*39.996 g/mole = 2.050 grams.
600524
Give the mass of one reagent in a reaction, and ask for the mass of another. A word equation or chemical equation must be provided.\n\neg:\n\nWhat mass of nitrogen gas is required to produce 45.0 g of ammonia, according to the reaction below ?\n\n         N2   + 3 H2   ---> 2 NH3\n\n\nTo solve the problem\n\na) convert mass of NH3 to moles (2.65 mol)\nb) use stoichiometry - find moles of N2 needed (1.32 mol) \nc) convert moles of N2 to mass (37.0 g)
835111
4 NH3  +  5O2  --->  4 NO  +  6H2O\n2 NO  +  O2 ---> 2 NO2  +  N2O4\n3 NO2  +  H2O ---> 2 HNO3  +  NO
774152
63.33 mL required.\n\nTo prepare the solution, you need 0.380 M * 2.00 L = 0.760 moles sulfuric acid.  To obtain 0.760 moles of sulfuric acid from a 12M sulfuric acid solution, you need 0.06333 liters, or 63.33 mL.
867776
Moles = Mass divided by molar mass\n\nMolar mass of (NH4)3CO3 is 114.16\n\nMoles = 0.795 / 114.16\n\n0.006964 moles
964051
First you must know the equation of the reaction, so you can establish the mole connections between the involved substances\n\n2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(l)\n\nNow, you have 60,0 mL of 0,100 M Mg(OH)2(aq)\nTherefore:\n0,100M*(60,0 mL)(1L/1000 mL)= 6,00*10^-3 moles of pure Mg(OH)2\nNow, 2 moles of HNO3(aq) react with 1 mol of Mg(OH)2(aq)\n\n6,00*10^-3 moles Mg(OH)2*(2 moles HNO3/1mol Mg(OH)2 )= 0,0120 moles of HNO3(aq)\nMolarity = moles solute/volume of solution\n0,200M= 0,0120 moles /?\n?= 2,40*10^-3 L of solution= \n2,40 mL of 0.200M HNO3(aq)  are needed to completely neutralize 60.0 ml of 0.100 M Mg(OH)2 solution
476691
As long as you are not changing the temperature or the number of moles of gas in the system, the following equation applies:\n\nP1V1 = P2V2\n\nthus (400 mL)*(500 torr) = (X mL)*(760 torr)\n\nSolve for X.
699263
V=nRT/P\nn(O2)=3/2(0.0022mol of KClO3) =0.0033 gmol\nR=0.0826\nT=294 k\np = p (O2) + p (water vapour)\np(O2) = 0.926 Atm\nV=(0.0033)(0.0826)(294)/0.926=0.087 liter = 87 ml
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
663581
NH4NO2 --------------> N2 + 2H20 \n\nYOU NEED TO HEAT THE SOLUTION IN ORDER TO DECOMPOSE THE COMPOUND \n \nHOPE THIS HELPS
48128
d = density\nM = molar mass or mass per mol\nV = volume\nn = number of mols\nR = gas constant\nT = temperature\nP = pressure\n\nd=nM/V \n\nV=nRT/P\n\nd=nMP/nRT\n\nd= MP/RT \n= (28x10^-3 kg/mol x 300x10^3 N/m^2)/(8.314 J/(mol K)x450K) \n\n= 2.25 kg/m^3
700367
1) Balance the reaction equation:\nPb(NO3)2  + 2 LiBr -> PbBr2 + 2LiNO3\n\n2) Calculate the amounts we start with\nPb(NO3)2 :  2.05 L * 0.0550 mol/L = 0.1128 mol\nLiBr : 1.52 L * 0.0618 mol/L = 0.0939 mol\nAs we need 2 moles of the latter for 1 mole of the former, the reaction is limited by the amount of LiBr\n\n3) Calculate the amount of PbBr2 produced\n0.0939 mol of LiBr will react with 0.470 mol of Pb(NO3)2 to produce 0.0470 mol of PbBr2. This is 0.0470 * (207.2 + 2* 79.9) =17.249 gram\n\n4) Calculate the yield\nThe yield is 15.784 / 17.249 = 91.51%
592234
3g of hidrogen (plus 14 g of nitrogen) make up 17g of ammonia (H+H+H+N=1+1+1+14), so:\n\n3g  H____17g NH3\n50g H____=283.33g NH3\n\nI assumed a 100% efficient reaction (that means every bit of reactant will actually react).
600936
Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O\n\nno. of mol of Na2CO3 reacted\n= m/M\n= 1.5/106.2\n= 0.01412 mol\n\nno. of mol of HCl reacted\n= 95/1000 x 6\n= 0.57 mol\n\nsince Na2CO3 is the limiting reagent, the no. of mol of CO2 liberated = 0.01412 mol\n\nMass of CO2 liberated\n= 0.01412 x 44.01\n= 0.6214g\n\nVol. of CO2 gas \n= 0.6214 x 1.96\n= 1.217\n= 1.22g (3s.f.)