How many moles of   hydrogen   were in the original sample that underwent combustion ?
969783
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 6.406 g x (12 / 44) = 1.747 g\nhydrogen: 5.244 g x (2 / 18) = 0.583 g\noxygen: 4.664 - 1.747 - .583 = 2.334 g\n\n2) convert to moles.\ncarbon: 1.747 / 12 = 0.146 mol\nhydrogen: 0.583 / 1 = 0.583 mol\noxygen: 2.334 / 16 = 0.146 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.146 / 0.146 = 1\nhydrogen: 0.583 / 0.146 = 4\noxygen: 0.146 / 0.146 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is CH4O.\nThe atomic masses add up to 32, which is also the molar mass of the compound.
272349
Ca(s) + 2HCl(aq)-->CaCl2 + H2(g)\n\nor...\n\nCalcium Chloride and Hydrogen Gas... you know... the stuff that fuels rockets! :)
914957
They did experiments where they were trying to see how much heat it takes for a fire to ignite from a disposable lighter.  If you concentrated a blow torch near the lighter, it exploded.  The lighter exploded after 400 degrees in a toaster oven.  For the final test, they stuck 500 lighters in a car, then put a heating element in the car to represent the car sitting on a hot day.  The car exploded.
293861
The combustion of glucose is as follows:\n\nC6H12O6 + 6O2 --> 6CO2 + 6H2O\n\nSo, the mole ratio is 1 mol glucose per 6 mols O2.
1116694
the reaction leads to production of hydrogen gas ...\n\n\n2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl-(aq) + 3H2(g)
856297
V=55.0L\nP[H2 = 2.00 atm\nP[O2 = 3.00 atm\nT[1 = 25°C = 298 K\nT[2 = 125°C = 398 K\nR = 0.082057 L · atm / (K · mol)\n\nIdeal gas law: PV = nRT\n\nAt 25°C:\nsolving n for H2:\nn = PV/RT = 2.00 * 55 / (0.082057 * 298) mol = 4.498 mol\n\nsolving n for O2:\nn = PV/RT = 3.00 * 55 / (0.082057 * 298) mol = 6.748 mol\n\nAt 125°C:\nsolving n for H2:\nn = PV/RT = 2.00 * 55 / (0.082057 * 398) mol = 3.368 mol\n\nsolving n for O2:\nn = PV/RT = 3.00 * 55 / (0.082057 * 398) mol = 5.052 mol\n\nThe Reaction that occurs is:\nH2 + 1/2 O2 = H2O\n\nSo, hydrogen is the limiting reactant, leaving water and oxygen as the resulting composition.\n\nAt 25°C, 4.498 mol of H2 and 2.249 mol of O2 form 4.498 mol of H2O, leaving 4.499 mol of O2.  The total number of moles (n) is 8.997 mol\n\nAt 125°C, 3.368 mol of H2 and 1.644 mol of O2 form 3.368 mol of H2O, leaving 3.408 mol of O2.  The total number of moles (n) is 6.776 mol\n\nUsing the ideal gas law again:\nPV=nRT\n\nP = nRT/V\n\nAt 25°C:\nP = 8.997 mol * 0.082057 L · atm / (K · mol) * 298 K / 55 L = 4.00 atm\n\nAt 125°C:\nP = 6.776 mol * 0.082057 L · atm / (K · mol) * 398 K / 55L = 3.00 atm
801336
If you neglect the loss in mass due to the energy given off in the process (a very, very, tiny amount of mass at that), then the total mass should stay constant before and after the reaction.\n\nIf you have the reaction,\nA --> B\nThe mass of A equals the mass of B\n\nHowever, that is not to say that what is leftover after burning has the same mass as the original substance burned.\nIn a standard combustion reaction, two products are produced, Carbon Dioxide and water.  Both of these products are in the gas phase and "float" away, taking with them part of the mass of the original substance.  If one could trap and collect all the particles of a reaction in an enclosed space, then the mass will be constant, but in an open system where things can escape, you will generally loose part of the mass in this manner.
952068
a) 0.01 mol\n\n1 mole of gas, at standard tremperature and pressure occupies 22.4 l.  You have exactly 1/100 of that, therefore 1/100 of a mole.
7636
maby the thoaory of molecular reaction particles in a astrnomical burn of 123.8657 cm per inshes aka.a volcano made of clay with vinager and baking soda inside
969093
H2 = 2g/mol   2.05g = 1.025mol\nCO = 28g/mol    15.85g = .57mol\n\n2H2 + CO --> CH3OH\n\nmust be reduced to \nH2 + (1/2)CO --> (1/2)CH3OH\n\n1.025mol + (1/2).57mol --> the quantity of the limiting reactant which is .285mol CH3OH which at 32g/mol = 9.12g
765956
A) 1\nB)2 moles of gas. ,1 mole of gas takes up 22.4 liters, so 44.8 L
912712
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 4.793 g x (12 / 44) = 1.307 g\nhydrogen: 2.452 g x (2 / 18) = 0.272 g\noxygen: 2.018 - 1.307 - .272 = 0.439 g\n\n2) convert to moles.\ncarbon: 1.569 / 12 = 0.109 mol\nhydrogen: 0.272 / 1 = 0.272 mol\noxygen: 0.439 / 16 = 0.0274 mol\n\n3) divide each molar amount by the lowest value to find small integers:\ncarbon: 0.109 / 0.0274 = 4\nhydrogen: 0.272  / 0.0274 = 10\noxygen: 0.0274 / 0.0274 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
249344
If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.\n\nWe solve this problem by first writing the equation\nN2 + 3H2 = 2NH3\nN2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g\nso 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent.
842808
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.847 g x (12 / 44) = 2.686 g\nhydrogen: 5.038 g x (2 / 18) = 0.560 g\noxygen: 4.146 - 2.686 - .560 = 0.900 g\n\n2) convert to moles.\ncarbon: 2.686 / 12 = 0.224 mol\nhydrogen: 0.560 / 1 = 0.560 mol\noxygen: 0.900 / 16 = 0.056 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.224 / 0.056 = 4\nhydrogen: 0.560 / 0.056 = 10\noxygen: 0.056 / 0.056 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
477943
you need 2 moles of hydrogen for each mole of oxygen so for four oxygen moles you need eight hydrogens. oxygen is the limiting reactant and you can make four water moles
853370
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.
327281
Water is created by burning hydrogen.