How many moles of oxygen are consumed each time.?
924669
We need the source of phosphorous for this equation.  Without it, we can't help you.
754271
2H2 + O2 --> 2H2O\nPV = nRT\nP = 2 + 3 = 5\nT = 25 + 273 = 298\n5.00 atm x 55.0 L = n x .08206 x 298\nn = 11.25 mol H2 and O2 mixture\n11.25 mol x (2 mol H2O / 3 mol) = 7.50 mol H2O\n\nP x 55.0 = 7.50 x .08206 x 298\nP = 3.33 atm
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
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
776910
1 mole of an ideal gas occupies 22.4 liters, so multiply that by your number of moles and you would have the total volume occupied by this amount of gas at STP.
576829
Answer is 1.41 moles. \n\nMass of an Oxygen molecule is 16 AMU (Atomic Mass Units) x 2 = 32 AMU (Since an Oxygen molecule has 2 Oxygen atoms)\n\nSo 32 grams of Oxygen gas will make 1 mole of it. Now 45 litres of Oxygen gas can be considered to be equivalent to 45 grams of it at STP.\n\nHence,\n\n32 grams of Oxygen gas is number of moles = 1\n\n45 grams of Oxygen gas is number of moles = (1/32) x 45 = 1.41 moles
843713
You need to first balance this equation:\n\nCH4 + 2O2 ---> CO2 + 2H2O\n\n# of mole of 80.2 g CH4 = 80.2 / (12 + 4) = 5.01 moles\n\n# of moles of O2 needed = 5.01 * 2 = 10.03 moles\n\nTotal mass of O2 = 10.03 * (16 * 2) = 320.8 g
121351
Are you tryn to have us do your hw!?!?\n\nThe eqn should be in your book.  I dont feel like digging mine out, but its something like;\n\nCp*g*(T2-T1)=energy\n\nSo what u want to find is (t2-t1) Ima call it x\n\n400*50*(x)=800  do the math.  I think thats it.\n\nfor the 2nd question, this one is easy.  The combustion is energy is for 1 mol (per mol).  If you only have .2 moles, the combustion amount is .2* 1370 .
840588
First of all, you need a chemical equation,\n\nNO + O2 ----> NO2\n\nThen, balance it,\n\nNO + 1/2 O2 -----> NO2\n\nNow, what this balanced equation tells you is that the mole ratio between NO and O2 is 2:1. i.e., you need 2 parts of NO to react with every 1 part of O2.\n\nNow, you need to calculate the total number of moles of NO and O2 respectively,\n\n# mole of NO = 17.26/(14+16) = 0.575 mole\n\n# mole of O2 = 13.15/(16+16) = 0.411 mole\n\nAs said before, for every 1 mole of O2, it takes 2 moles of NO to react. From the number of moles of NO and O2, it is clear to see that NO is the limiting reagent. Why? Because if you react all 0.575 mole of NO, it takes only 0.575/2 mole (0.2875 mole) of O2, which you have a lot more than that.\n\nThe limiting reagent is the substance that is NOT in excess, which then determines how much product will be produced (because once this substance is all gone, there will be no reactants to react and give product).\n\nNow, from the balanced equation, you know the mole ratio between NO and NO2 is 1:1. Therefore, the number of moles of NO2 that be formed will be 0.573 moles.\n\nMass of NO2 = # mole * molecular weight\n= 0.573 * (14 + 16 + 16) = 26.465 g
756159
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l
796693
4PH3 (g)  ---> P4(g) + 6H2(g)\n\n4 moles of PH3 are consumed then 1 mole of P4 is produced and 6 moles of H2 are produced.\n\nIf you are consuming 0.0048 mol/s of PH3, then by the relation above you will be producing:\n\n0.0048/4 = 0.0012 mol/s P4\n0.0048*6/4 = 0.072 mol/s H2.
781068
The balanced equation for the reaction is:\n\n2 H2O2 --> 2 H2O + O2\n\n1]  If one mole of Oxygen is produced, the 2 moles of Hydrogen peroxide would be needed.  One mole of Hydrogen peroxide contains 2*H + 2*O = 2(1) + 2(16) = 2 + 32 = 34 grams.\n\n2 moles of Hydrogem peroxide = 2*34 grams = 68 grams.\n\n2]  Is it 0.3 moles or 0.3 cubic decimeters of Hydrogen peroxide?\n\nAssuming 0.3 moles, then from the above, 0.15 moles of Oxygen would be produced.  The volume would be:\n\n0.15 moles * 24 cubic dm/mole  = 3.6 cubic dm of Oxygen at rtp.
751855
The first step is to balance the equation first, which yields\n\n2SO2 + O2 ---> 2SO3\n\nNow, find the total number of moles of 160 g O2:\n\n# of mole = 160/(16*2)\n= 5 moles.\n\nAccording to the balanced equation, the mole ratio between SO2 and O2 is 2:1. In other words, it takes twice amount (in moles) of SO2 to react with 1 unit of O2.\n\nSince we react 5 moles of O2, we need 10 moles of SO2.\n\nMass of SO2 = # mole * molecular weight\n= 10 * (32 + 16*2)\n= 640 g of SO2 is needed.
825261
First you must balance the equation:\n4NH3 + 5O2  -> 4NO + 6H2O\nthere are 4N, 12H and 10O on each side\nyou have 4.78 moles of O2 you need to divide it by the number of moles of O2 that react and multiply it by the number of moles of NO produced.\n4.78 / 5 x 4 = 3.824moles\nHope this helps
316922
Ok.... \nfirst: 1atm = 760 torr\n Now, as per the gaseous  STP conditions., 1 mole of oxygen will occupy 22.4 lit volume at STP conditions.\nSTP: T=273K\n        P= 760 torr\n       V =22.4 lit\n       Mass of O2 at STP = 32g\n\nNow. second set of conditons:\n P2= 1400 torr\nT2 =41 K\nMass= ?\n\nPV = nRT\nPV = (w/Mwt) XRT\n\nP1/w1T1 = (P2 /W2 T2)\n\nSubstitute these values;;\nW2= 392g of oxygen.\n\nIs the solution is clear enough?
702103
1) Determine how many moles of C3H8 we have:\n147 g/ (3*12.01+8*1.008) =3.33 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nC3H8 + 5 O2 -> 3 CO2 + 4 H2O \nWe need 5*3.33 = 16.67 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. Determine at the given conditions:\nV(1mol) = 22.4 * (303/273) = 24.86 liter \n\nTherefore 16.67 mol = 414 liter = 0.414 m^3
795154
2CO(g) + O2(g) --> 2CO2(g)