If 9.988g H2 combines with N2, how many g product will be formed?
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?
1062523
2NaI + Hg(NO3)2 --> HgI2 + NaNO3\nwe have;\n52 ml and a 0.78M Hg(NO3)\ntherefore we can find # of Moles\nn = CV\nn = 0.78 * 0.052                       Concentration should be in Litres\nn = 0.04056 mol\n\naccording to the equation;\n2 moles of NaI react with 1 mole of Hg(NO3)2\ntherefore 2(0.04056) moles of NaI needed in the reaction\ni.e. 0.08112\nwe know V i.e. Volume equals the ratio of #of moles and Concentration \ntherefore ; \nV= n/C\nV= 0.08112 / 1.60\nV= 0.0507 L\ni.e. 50.7 ml of NaI required in the reaction !!
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
926891
just count how many of each element you have, and make sure they add up\n\n\nC2H6 + 5O2 ---> 2CO2 + 3H2O\n2 carbon 6 hydrogen 2 oxygen----> 1 carbon 2 oxygen 2 hydrogen 1 oxygen\n\n\nso to even it out it would be...\n\nC2H6 + 7O2--->2CO2 + 3H2O\n2 carbon 6 hydrogen + 7oxygen--->2 carbon 4 oxygen + 6 hydrogen 3 oxygen\n\niff you add up each element they are the same on each side
187568
You can come up with a molecular weight of air - use the percentages of the different molecules comprising air and the molecular weight of the components.  The answer is around 29 g/mol (I think).
753121
Ok, now this one I can do:\n\nMg3N2  ---> 3Mg (s) + N2 (g)
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
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
836906
I had to read about this in order to answer your question at this web site:\n\nhttp://home.um.edu.mt/chemistry/CH230/2.Analytical/CP210a.PDF\n\nSo, two moles of DMG combines with each Ni++ ion.\n\n(0.2145 g Ni)* (1 mole Ni/58.71 g Ni) = 3.65*10^-3 mole Ni\n\nIf what you tell me is correct that DMG is 116.1194 then the Ni complex which loses 2 H+ for the Ni++ has a molecular wt of\n172.8134 (please double check)\n\n(3.65*10^-3 mole Ni)* (172.8134 g/mole) = 0.6314 g\n\nplease check for errors
929547
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
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).
878226
The method is very simple.  I can do the first and you can try the others.\n\n1.  First write the reaction and balance it.\n\n2HgO ---> 2Hg + O2.\n\n2. Convert grams to moles of whatever mass you have.  In this case you have O2 as 96 grams.  You convert to moles by dividing the mass by the molecular weight.  In this case, for O2 will be 32 g/mol\n\n96 grams / 32 g/mol = 3 mol.\n\n3. Now you look back to the equation.  In the equation for producing 1 mol O2 you need 2 moles of HgO.  If now have 3 moles O2, then you will need 2 x 3 moles of HgO, 6 moles.\n\n4. Now you need to convert the moles of HgO to grams.  To do this calculate the molecular weight of HgO.  It is 216.6 g/mol\n\n6 moles * 216.6 g/mol = 1299.6 g of HgO.
693257
From your reaction equation, 3 moles of hydrogen gas produces 2 moles of ammonia if the reaction was 100% efficient.  From the way you phrased your question, it's difficult to tell if you mean 450g of hydrogen (H), or 450 grams of hydrogen gas (H2).  I'll assume you mean hydrogen gas.\n\nFirst you have to calculate the number of moles hydrogen gas you started with:\n\nMoles H2 = grams H2 / Molecular weight H2\nMoles H2= 450 g H2 / 2.016 g per mole H2\nMoles H2 =223.21 moles\n\nIf the reaction was 100% efficient:\nMoles NH3 = 2/3 * moles H2\nMoles NH3 = 2/3 * 223.21\nMoles NH3 = 148.80 moles \n\nNow to calculate the grams NH3 expected from 100% efficiency:\ngrams NH3 = moles NH3 at 100% efficiency * mol wt NH3\ngrams NH3 = 148.80 moles * 17.024 grams/mole\ngrams NH3 = 2533.17 grams\n\nSo to calculate the percent yield for the reaction:\n\n% yield = (actual yield/ theoretical yield) *100\n% yield = (1575 grams/ 2533.17 grams) *100\n% yield = 62.17%\n\nNow if you meant H instead of H2 for the original grams hydrogen in your question, you'll have to recalculate as appropriate.
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
739541
CH4 + 2O2-> CO2 + 2H2O
855371
Look at like this. Whatever the amount of the first component\nis, consider it 100%, then for the second component, it would be 80% of the first.\nExample:\nYou have 500 gm of component 1,\nyou put 80%of that or 400 gm of component 2.\nThe total gms would be 900 gm, 100 parts component 1 to\n80 parts component 2.\nHope this helps.
926935
PV = nRT\n\n685 mmHg / 760 = .901 atmospheres\n\n18.6 mmHg / 760 = .0245 atm\n\n.901-.0245 = .8765 atm\n\n21 degrees c = 294 K\n\n.8765 atm x 7.80 L = n x .08206 x 294 K\n\nsolve for n then use the molar mass of H2 (2.016) to convert to grams.\n\nmy calculator's out in the car or else i'd do it for you.