How many liter of H2 would react with 4 liters of N2?
917667
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
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).
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.
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
1101057
2N2H4 + N2O4 -> 3N2 + 4H2O\n\nn(N2H4) = 3.125 mol\nn(N2O4) = 3.333 mol\n\nAs 2mol N2H4 rx c 1mol N2O4, n(N2H4rxn) = 1.563mol, and hence it is the limiting reactant. n(N2O4) = n(N2) = 1:3\nTherefore, 1.563 * 3 = 4.688 mol of Nitrogen gas, which equates to 131.25g.
766259
4 NH3 + 3 O2 ---> 2 N2 + 6 H20
384394
Well, let's see:  you're releasing 1000 kg/hr of liquid N2, which equilibrates to 1 atmosphere and 20C (STP).\n\nAt STP, one mole of a gas occupies 22.4 liters;\nyou're releasing 1000 kg/hr = (1000 kg*1000g/kg)/28g/mole = 35,714 moles of N2 per hour, and thus:\n\n22.4 liters/mole * 35,714 moles = 800,000 liters (at STP) per hour
638965
What is "Dry N2 gas?"  My chemistry's a bit rusty, but here's what you might do:\n\nYou take your 40 liters and with the 22C and .956atm, you can determine how many moles of N2 you have.\n\nThen, you take those moles and apply it to whatever "Dry N2 gas" means.  My best guess would be STP.  Then you find the volume at the new temperature and pressure.
813197
0.0821 moleP1V1/T1=P2V2/T2\n375*4/293=760*V2/273.5\nV2=375*4*273.5/(293*760)\n22.4 litres contains 1mole\n375*4*273/293*760\n=375*4*273/293*760*22.4\n=0.0821 mole
433864
Use the General Gas Equation:\n\nPV = nRT\n\nwhere\nP = pressure \nV = volume \nn = number of moles\nR = Universal Gas Constant (value depends on units of P,V, n, T eg R = 62.3638 when V is in litres, P is in mmHg, T in Kelvins and n in moles.)\n\nUse the above equation to solve for n. If you want the answer in grams, simply multiply the number of moles by the molar mass of H2.\n\nIf you require a more precise answer, then you should use the Van der Waals equation (a more sophisticated version of the General Gas Equation that takes into account interactions between particles and the volume of those particles).
453509
i'm not familiar with the unit barg, but just use the ideal gas equation PV=nRT solved for n so n=(PV)/(RT) with an apropriate value for the universal gas constant (R).  that will give you the number of moles of N2 which you can then determine the volume under standard conditions if you wish using the same equation.
806996
the moles of O2  : 4/32 = 0.125\nthe volume occupied by O2 at STP : V = n*22.4 = 0.125 * 22.4 =2.8L
1186593
Mix with a 0% solution in the relation 33:2\n\nSo use 2 units of your solution (like 2 liters) and 33 units of the 0% solution (like 33 liters).\n\nThat's all.
739541
CH4 + 2O2-> CO2 + 2H2O
315401
Use PV=nRT ideal gas equation.\nP=0.400 atm\nV=?\nn=3 mol\nR=0.0821 L atm /K/Mol\nT=300K\n\nDo yourself now...
552992
ch4 for both.\nOne carbon atom and 4 hydrogen atoms. Simple!
720117
5.  K = [NH3][HCl] = (0.066 M)^2 = 4.36 x 10^-3\n\n9.  I'm assuming that equation should be H2(g) + CO2(g) <---> H2O(g) + CO (g)\na.  Add H2O will drive equlibrium to the left\nb.  Removing CO will drive equilibrium to the right\nc.  Adding H2 will drive equilibrium to the right\nd.  Adding something to absorb CO2 will drive equilibrium to the left\n\n6. \na.  Increasing the volume of the reaction vessel will reduce the pressure, favoring the production of more gaseous products, so the equilibrium will go to the right\nb.  Equal number of moles of gas on both sides, so increasing the volume will have no effect.\n\n1.  \na.  Decreasing the volume increases the pressure, favoring the side with fewer moles of gas.  Equilibrium would go to the left.\nb.  Equilibrium would go to the left, as there are fewer moles of gas on that side.