H2 is the limiting reagent so there is excess N2? \n\nThanks for your help!!
203920
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?
180049
When performing a chemical reaction you want to use as little of each material to generate the quantity of material you need, with as little waste as possible.\n\nFor example:\n\nHCl + NaOH -> NaCl + H2O\n\nYou are using 1 mole of HCl and 1 mole of NaOH to generate 1 mole of salt and 1 mole of water, therefore stoichiometryt is 1:1.  This is the stoichiometry.  \n\nThe reaction:\n\nH2SO4 + 2NaOH -> Na2SO4 + 2H2O\n\nThe stoiciometry is 1 mole of H2SO4 and 2 moles of NaOH, therefore 1:2\n\nStoichiometry is the relative amounts of each reagent you use in a reaction.\n\nIf you were to use:\n\n2HCl + NaOH -> NaCl + H2O + HCl\n\nYou are using 2 moles of HCl and 1 mole of NaOH to generate the same thing, but you have an excess of HCl left over.  This is yield.  The yield of this experiment in relation to HCl would be 50%, since you have only consumed half of it, the rest is waste.  With respect to NaOH the yield is 100% since you have used all of it.  The yield of the reaction though is 100% as it is always based on the limiting reagent.\n\nThe product yield is the amount of product you recover based on a theoretical maximum.  In the first example above, and assuming you did the reaction on a 1 mole scale, you should recover 1 mole of NaCl (57.5g) and 1 mole of H2O (18g).  If you only recovered 30g of NaCl, then the yield of the reaction is 30 / 57.5 = 52.2%.\n\nNo chemical reaction is perfect so getting 100% return is rarely possible, although we do get very close occasionally in industry.\n\nEcologically the less EXCESS of reagent you can put into a reaction to maximise the recovery limits the amount of material going to waste, which has an impact on the environment.  Economically the less EXCESS of reagent you need, means it costs less to prepare a compound.\n\nWhen pharmaceutical companies make drugs they spend millions on materials.  Cutting down by 10% how much of everything they need can save hundreds of thousands of pounds per year / per drug on excess materials expenditure, increased environmental exposure etc.  Chemical companies spend 100's of thousands of pounds a year disposing of excess chemicals safely, recycling chemicals etc and are liable to very heavy fines if they don't.  Many savings can be made just by improving stoichiometries and product yields.\n\nLast year I spent over $1 million on chemicals, by improving stoichiometries and yields by 10% this alone saves an enormous amount of money.\n\nI hope this helps?
246053
This is a test of the basic principle of partial pressures.  The pressure of each gas is independent of the other gasses.  If you fill a 2 liter container with Neon to 2 atm, the neon will exert 2 atm no matter what other gas you put in the container.  This means that if the N2 originally exerted 4 atm, it will exert a 4 atm partial pressure when combined with the other gasses and the overall pressure will be 2+3+4 = 9 atm.
473444
The first reaction is simply wrong.  Nitrous acid does not ionize to nitrate.\n\nThe second reaction is a precipitation reaction, showing the net ionic equation.  The compound is ionic.\n\nThe third reaction is an example of SN2 displacement, not elimination.  Only single bonds are formed.
536609
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
338252
It's not nitrogen... It's hydrogen. They called my car the "Hindenburg" after it burst into flames when I pumped in a few pounds too many way back when in Lakehurst, New Jersey... Oh, the humanity!
58999
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
384105
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...
456132
The concept of 1.5M hydrogen gas is rather odd, but, yes, you use the Nernst Equation.  Don't forget to convert the pH to a molarity of H+ ions.
533459
2NO + O2 --> 2NO2\nMr of NO = 14 + 16 = 30\nMoles of NO = (29.03/30) = 0.9677 (to 4 s.f.)\nMr of O2 = 32\nMoles of O2 = (19.35/32) = 0.6047 (to 4 s.f.)\n\nComparing the moles of NO and O2 given, you should notice that NO is the limiting reagent.\nHence moles of NO2 produced = moles of NO used = 0.9677\n(since the ratio of NO2 to NO is 2:2)\nMass of NO2 produced = 0.9677 x (Mr of NO2) = 0.9677 x 46 = 44.5 g (to 3 s.f.)\n\nHope this helps :)
419467
ooo this will be fun!\n\nok..\nA)\n2.207g of H2O MW. 18.0\nmol = weight of sample (g) / molar weight (g/mol) (n=m/M)\nn = 2.207 / 18\nn = 0.1226 moles of H2O\n\nnow, since there is 2 hydrogens in H2O, then when the molecule is broken down, there is twice as many moles of hydrogen then the overall molecule,\ntherefore,\n\n2x 0.1226 = 0.2452 moles of H.\n\ndetermining number of moles of CO2 is also helpful,\n4.313g, n = 4.313 / (12.01 + 16 +16) = 0.0980moles\napprox, 5(moles H2O) :  4(moles CO2)\n\nB) \nworking the number of moles of this compound will be useful:\nn = 1.816 / 74.122 = 0.0245 moles, which is 1/10 of the moles of H produced. (or 1:5 of the moles produced of water)\n\nif you think about it, the reaction is something like this:\nCH2 + O2 --> CO2 + H2O\nclearly the above doesn't make sense, because 1) it isn't balanced, and 2) the reactant needs a higher molecular wt.. now thinking about it, the following can be deteremined:\nand from the ratio of moles found above:\n       1     : (excess):  4        :    5\nC4H10 + 13/2O2 --> 4CO2 + 5H2O\n\nnow, I amoung most allow fractions in equations, but if you don't, multiply through by 2.\n2C4H10 + 13O2 --> 8CO2 + 10H2O\n\nnow, you want the empirical forumla,\ntherefore,\ndivide by highest common factor,\nC2H5 \n\nthe empircal fomula is just a ratio and doesn't make huge sense, as the formula for an alkane is CnH2n+2\nhope that makes sense :)
243020
here, you need to apply the concept of limiting reagent.\n\nhere, N2 is limiting reagent. \n\nSo, if 28 gm (molar mass of N2)nitrogen would give you 34 gm of (molar masss of 2NH3) ammonia.\n\nSo, 3 gm (given) nitrogen would give you how many?\n\n = (7 * 34)/ 28\n\n = 8.5 gm of ammonia.
280979
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).
212887
(a) 1,2-dibromobut-1-ene\n(b) 2,2-Dichlorobutane (based on Markovnikovs rule)\n(c) don't know\n(d) don't know.  Na is often used to reduce multiple bonds in a free-radical manner, however you need a source of H.\n(e) KMnO4 is a very strong oxidising agent and is commonly used as tests for unsaturation.  The side product is MnO2, showing that two atoms of oxygen are transferred to the product.  The product formed I think is 1-hydroxy-butan-2-one.
526446
H2S + 2NaOH --> Na2S + 2H2O\nMoles of H2S given = (1.19/34) = 0.035\nMoles of NaOH given = (4.91/40) = 0.12275\nFrom the chemical equation, 1 mole of H2S needs 2 moles of NaOH for complete reaction. Hence for 0.035 moles of H2S, you need only (0.035 x 2 = 0.07) moles of NaOH.\nThus the limiting reagent is H2S.\nFrom chemical equation, 1 mole of H2S produces 1 mole of Na2S. Thus moles of Na2S produced = moles of H2S used = 0.035\nMass of 0.035 mole of Na2S = mole x Mr of Na2S = 0.035 x 78 = 2.73g\n\nHope this helps :)
444154
The equation doesn't balance.  There's no hydrogen on the right side.  But, assuming that you ended up with H2 gas on the right (I don't think that reaction works real-world, but whatever), the main thing you need to look at is the chlorine.  It's Cl2 on one side and Cl3 on the other.  That means that you must have a total number of chlorine atoms which is a multiple of 6.  Try balancing the equation with 6 chlorine atoms and see what you get:\n3CuCl2 + 3H2O + 2Al -----> 2AlCl3 + 3 CuO + 3H2
365451
Use the limiting reactant.