How many grams of oxygen are needed to burn a gallon of gas?
914564
There is about 20% oxygen in the air which is more than enough.\n\nAlthough a pound is 454 grams. you asked about gallons. Gasoline is about 6 pounds per gallon.\n\nMost gasoline is burned in internal combustion engines which do not burn it completely.It forms carbon monoxide.\n\nGasoline is not pure octane, C8H18 \n\n 2C8H18 + 25O2 ---> 16CO2 + 18H2O \n\nNow you can work out the rest, 2 moles of octane to 25 moles of O2. Use themolecular weight per mole of each.Figure out how many moles of octane yoy have.
558286
You are completely right, the amount of oxygen has indeed been reduced through the burning of fossil fuels. Reference attached to the Columbia University web page of the world's foremost authority on the global oxygen cycle.\n\nWhile no danger exists that our O2 reserve will be depleted, nevertheless the O2 content of our atmosphere is slowly declining--so slowly that a sufficiently accurate technique to measure this change wasn't developed until the late 1980s. Ralph Keeling, its developer, showed that between 1989 and 1994 the O2 content of the atmosphere decreased at an average annual rate of 2 parts per million. Considering that the atmosphere contains 210,000 parts per million, one can see why this measurement proved so difficult.
365485
I assume you mean molecular weight.\n\nSince carbon weights 12 grams, and oxygen weighs 16:\n12+(16*2) = 44 grams.\n\nYou have to give the amount of CO^2, in ML to get Moles.\n\nSo generically speaking, 1 mol of CO^2 weighs 44 grams, 2 mol weights 88 grams, and so on.\n\nTo calculate the amount of atoms in it, you would have to times it with Avogadro's number. (6.0221415 × 10)
865243
C3H6 +  9/2 O2 ---> 3 CO2 + 3 H2O\nassumed oxygen is ideal gas, so\nPV=nRT\nR = 82.057 cm3.atm/gmol.K\nV = 15 l = 15000 cm3 ; T = 37 C = 310 K\nn (mol) O2 = PV/RT = 2.4*15000/82.057*310 = 1.415 gmol\nfrom balanced reaction CO2 = 3/(9/2) * 1.415 gmol = 0.943 gmol\n                                         = 0.943 * 44 = 41.507 gram
824604
To end up with exactly one gallon, you would need 124.12 ounces to 3.87 ounces
714204
What teacher still gives exercises in such mixed units? Degrees Fahrenheit and Kelvins. Distance in feet....but the results is required in grams?\n\nThe volume of air is 16000 ft^3 or 453 m^3. The density of air is 1.293 gram/liter (I looked that up) or 1293 gram/m^3. We have therefore 5.86 x 10^5 grams of air to heat.\nThe temperature change is 30 deg F or 16.7 deg K.\nSo the total heat needed is 1.01 x 16.7 x 5.86 x 10^5 -9.88 x 10^7 Joule. \nThe combustion of methane provides 890.8 x 10^3 J/mol, which means that we need 111 mol of methane or 111 x (12+4x1) = 1775 gram of CH4\n\nThe reaction equation is CH4 + 2 O2 -> CO2 + 2 H2O.\nSo one mol of methane will produce one mol of CO2. A total of 111 mol are produced, or 111 x (12+2x16) = 4884 gram of CO2.\n\n(And all that stuff goes in the atmosphere warming up the greenhouse a little more.)
113809
It is true that at times, we need to know conversion factors and we just don't know the source.\nWell, if you are online, Google can come to the rescue.\n\n Just say you want to know that "1 gallon = ? ounces"\n\n Then just type the sentence in the inverted commas in the google search box at http://www.google.com and click on search. Instantly you will get the answer that:\n1 US gallon = 128 US fluid ounce\nIf you find trouble doing so, check this link below:\n\nhttp://www.google.co.in/search?hs=OKD&hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&q=1+gallon+%3D+%3F+ounce&btnG=Search&meta=\n\nI hope that you understand how to find conversions using Google now. That's more important.
147861
Fires need oxygen to burn.
569208
You need the specific heat of water, which is 4.184 J/g*C.  Assuming a density of water of 1 g/ml (not always the safest assumption, but close enough), then your 500 ml = 500 g.  You're raising the temperature 35 degrees, so the heat you will need will be:\n\n4.184 * 500 * 35 = 73,220 J = 73.22 kJ.\n\nThe combustion of decane provides 6.78 MJ/mol, or 6780 kJ/mol.  In order to give 73.22 kJ, you would need to burn 73.22/6780 = 0.0108 mol of decane.  Given decane's molecular weight of 132 g/mol, you would have to burn 1.425 grams.
714714
A balanced equation for this combustion is:\n\nC3H8 + 5 O2 ----> 3 CO2 + 4 H2O\n\nSo, every mole of C3H8 requires 5 moles of O2 to burn.\n\n# of moles of C3H8 = 138 / (12*3+8) =3.136 mole\n\n# of moles of O2 required = 3.136*5 = 15.682 moles\n\nUsing ideal gas law, PV = nRT, we get\n\nV = nRT / P = 15.682*8.314*(25+273) / (100*1000)\nV = 0.389 m^3
780338
Oxygen = O\nNitrogen = N\nCarbon Dioxide = CO2\n\nThey take up the space that is allowed them by the gravity and temperature of the planet, but oxygen and nitrogen form the main two constituents of the atmosphere.
363702
changes oxygen into carbon dioxide, the carbon come from the burning fuel
96968
mass of C4H10 = 48.0amu + 10.1amu = 58.1 amu\n(100.0g C4H10/1)(1 mole C4H10/58.1g)= 1.72 moles C4H10\nSo 100 grams of C4H10 is 1.72 moles.\n\n(1.72 moles C4H10/1)(13 moles O2/2 moles C4H10)= 11.2 moles 02\nSo you need 11.2 moles of oxygen to burn 100 grams of C4H10.\n\n(11.2 moles O2/1)(32.0g/1 mole O2)= 358g O2\n\nYou need 358 grams of oxygen.
765956
A) 1\nB)2 moles of gas. ,1 mole of gas takes up 22.4 liters, so 44.8 L
54895
To be precise, the answer is no. Do u noe why? The mean mol wt of air is 29 whereas that of CO is 28.\n\nlet me show u how i get these answers:\nC=12,O=16,mol wt of O2=32g/gmol,mol wt of N2=28g/gmol\nmol wt of CO = 12+16\n             =28\nas for air, by common sense, we know that air is made up of O2,N2, and other gases,rite? As the mol composition of other gases are very negligible(<1%), we assume air is only made up of O2 and N2. Air consists of 21mol% O2 and 79mol% N2.\n\nlet's take a basis of 100gmol(or 100mol)\nmoles of O2=21gmol\nmoles of N2=79gmol\nmass of O2=21X 32g/gmol\n          =672g\nmass of N2=79X28g/gmol\n          =2212g\ntotal mass= 2212+672\n          = 2884g\naverage mol wt of air=total mass/total moles\n                     =2884g/100gmol\n                     =28.84g/gmol\n                     =29g/gmol(whole no.)\n\nsince the ave mol wt of air is greater than that of CO, so aCO is not heavier than air,instead,slightly lighter than air.\n\nDoes it answer ur question?
369334
of course gas but if cant get that any flex fuel additive sold at auto stores. (or can order by gallon for you) with common ratio of 1 pint additive to 5-10 gals coco or other subsitute. theyll know which is best. also if use gas youd want 1 gal per10 gals of your substitute fuel.
977270
first you need a balanced equation.\n\n2Mg+O2===>2MgO\n\nNext convert everything given to mols.\n\n10.000 g Mg X 1 mol Mg / 24.30 g Mg (NOTE: this number is from the periodic table of the Elements) = 0.4115 mol Mg\n\n5.267 g O2 X 1 mol O2/ 32.00 g O2 (this is the molar mass of O2)= 0.1646 mol O2\n\nAt this point pick one of the mols you calculated.\n\n0.1646 mol O2 X 2 mol Mg (from balanced equation) / 1 mol O2 (from balanced equation)(for every 2 mol of Mg there is 1 mol of O2)= 0.3292 mol Mg needed to react with all of O2\n\nSince more Mg is available then needed (0.4115 mol > 0.3292 mol) this implies that O2 is the limiting factor. \n\nUse the limiting factor (O2) to get the mass of the MgO\n\n0.1646 mol O2 X 2 mol MgO / 1 mol O2= 0.3292 mol MgO\n\n0.3292 mol MgO X 40.30 g MgO/ 1 mol MgO = 13.27 g MgO\n\nessentially for all the other ones you are doing the same thing as before but with different numbers.