What volume (ml) of   O2   is collected over water at   24.0   oC   by the reaction of ..(see below)?
825631
Change grams to moles: 0.254 g/122.5g/mole = 0.00207 moles\n\nChange to moles O2 (ratio from equation is 3/2)\n0.00207 x 3/2 = 0.00311 moles O2\nCalculate vol. from relationship 22.4L/mole for any gas at STP\n22.4L/mole x 00311mole = 0.0697 L at STP\n\nUse Combined Gas Law to change to vol. at 24.0 degrees and 729.1mm. Reduce pressure by the water vapor pressure.\n\nP1 = 760.mm V1 = 0.0697 T1 = 0 + 273\nP2 = 729.1-22.4 V2 = x T2 = 24 + 273\n\nx = (0.0697)(760)(297)/(273)(706.7)... = 0.0815 L or 81.5 mL
11182
stochiometrically\nthe reaction is as under\n2H2 + O2 --> 2H2O\n\nhere 2 moles H2O is produced. Now to produce 2 moes of water 2 moles H2 is required. So to produce 8moles water 8 moles H2 will be required
973272
ok, first go to moles.\n\n24.0g LiOH (1mol/23.948g) = 1.0022mol\n\nnow get liters using molarity\n\n1.0022mol (1L/0.153mol) = 6.55L or 6,550mL \n\nRight, there ya go, remember that you have 3 sig.figs. so you may want to represent significance by using scientific notation. So...\n6.55x10^3mL\n\nIf you're ever stuck, just simply go to moles and that will get you started (and usually gets you some points on tests).
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
878297
Assuming 1 mole (28.01348 g) N2 = 22.4 liters at STP, I get (28.01348*.00220102/22.4) =  0.0027526 g if collected at STP. Correcting by 273/293 * (1.5*760-17.5)/760 (for temp., press. and water) = 0.003788 g. \nSo I basically agree with your results. What's the ans. supposed to be?
702491
Mn2+ +2NH3 +2H2O  <-->  Mn(OH)2 + 2NH4+ (+:sup, 4:sub)\n4Mn(OH)2 + O2 + 2H2O  <-->  4Mn(OH)3\n2Mn(OH)2 + O2  <-->  2(MnO2 . H2O)\n\nMn(OH)2 white sediment\nMn(OH)3 brown sediment\nThe oxidization is faster if we add H2O2
391116
2Ca(s) + O2(g) -> 2CaO(s)
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
733966
So...find the balanced equation.\n\n2NO + O2 = 2NO2\n\nFind the masses of NO and O2.\n\nNO= 14g + 16g = 30g\nO2 = 2(16g) = 32g\n\nFind out the # of moles of NO you have...\n\n1mol/30g NO x 22.58g NO = .7527mol NO\n\nFind out how many moles of O2 will react with NO\n\n1 mol O2/ 2 mol NO x .7527 NO = .3763 mol O2\n\nFind out grams of O2 that will actually react...\n\n.3763mol O2 x 32g/1 mol O2 = 12.0427g O2\n\nAdd the amount of NO with the amount of O2 that will react...\n\n22.58g NO + 12.0427g O2 = 34.623g NO2\n\nI am NOT sure about this answer, I haven't done this in a long time....... Hope it helps.
202815
100g O2 x (1 mole O2)/(32g O2) x (22.4 L O2 @ STP)/(1 mole O2 @ STP) x (1/1.5) x (298/273)\n\n= 50.9 L\n\nThe 22.4 L = 1 mole is only true @ STP.  The higher pressure reduces that volume and the higher temperature increases it.
792924
A.\n2*22.4 l.\n\nB.\n1*22.4 l.\n\nOne mole of any gas occupy 22.4 l. at STP.
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
807347
Since it takes 1 mole of O2 to react with H4 and get H2O + H2O. I would have to say that .264 moles of H2O were created. water is 18gper mole so .264*18g=4.752g of water.
777672
S2O3(2-)-->SO4(2-)          balance S first\nS2O3(2-)-->2SO4(2-)         balance O with water\n5H2O+S2O3(2-)-->2SO4(2-)        balance H with H+\n5H2O+S2O3(2-)-->10H++2SO4(2-)       balance charge with e-\n\n5H2O+S2O3(2-)-->10H++2SO4(2-)\n                                        +8e-\nthe 8e- is with the product side
1107016
To find amount of water produced, first , we need to write a balanced equation for the reaction:  \n\n2H2 + O2 = 2H2O\n\nTo do any Stoechiometry calculations we need to change the grams to moles.\n\n1 g H2 * 1 mol/2.016 g = .496 moles of H2\n10 g O2 * 1 mol /32 g =.3125 moles \n\nThe ratio between H2 and H2O in the balance equation is 1:1\nOn the other hand, the ratio between O2 and H2O in the balanced equation is 1:2.  \nThat means we can make .496 moles of H2O from the given H2 and 03125*2 = .625 moles of H2O from the given O2.  In this case H2 is the limiting reactant since H2 will run out first. That means we can only make .496 moles of H2O from the given H2 and O2. \nNow we need to change the moles of H2O to grams.  \n.496 moles * 18.016 g/1 mol H2O = 8.936 grams.
275919
2H20 --> 2H2 + O2\n\ntheoreticalyy spk it's exactly half the vol of H2, just look over the stoechiometrical balance above; but due to gas losses, it may be less
329297
decomposition reaction\n\n2H2O2 -------- O2   +  2H20