What volume of   O2 (m3)   is needed for the complete combustion of   160 g   of   C3H8   at   12.0oC   and?
961389
first, write down the reaction:\n 2C3H8 + 7O2 -> 6CO2 + 8H2O\n\nthen, convert gram of C3H8 to mol -> 160/44 = 3.64 mol.\nthird, find the number of mol needed to burn C3H8 by looking at the coefficient of the O2 and C3H8 in the reaction.\nn of O2 = 7/2 * 3.64 = 12.74 mol.\n\nafter getting mol of O2, we can compute the volume of O2 using:\nPV = nRT. remember that the temperature is in Kelvin not in Celcius. the pressure must also be in atm not in Pa.
760656
mole of O2 = 1 mole\nmole of CH4 needed = 0.5 mole\n\nBy using ideal gas theorem:\nP * V = n * R * T\n1.5 * V = 0.5 * 0.082 * 300\nV = 8.2 litre\n\npeace\nvixklen
49337
At room temperature, 0.968 g/cm³
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
228649
CO2 + H2O + light = C6H12O6 + O2
899695
The molecular weight of C2H6 is 30 g/mol, so 138 g = 138 g / 30 g/mol = 4.60 mol\n\nThe balanced combustion reaction of ethane is this:\nC2H6 + 3.5 O2 --> 2 CO2 + 3 H2O\n\nSo, for each mole of C2H6, you need 3.5 moles of O2, so you need 3.5 * 4.60 mol = 16.1 mol\n\nUse the ideal gas law to find the volume:\nPV = nRT\nP = 100 kPa\nn = 16.1 mol\nR = 0.008314 m3 kPa / (mol °K) : the ideal gas constant\nT = 27°C = 300°K\n\nV = nRT/P = 16.1 * 0.008314 * 300/100 m3\nV =  0.402 m3
739541
CH4 + 2O2-> CO2 + 2H2O
1170712
The short answer to your question is that the air to fuel ratio for 4 stroke petrol engines is 14:1 (by volume of gas (not gasoline!)). \n\nTo work it out from first principles, you need the equation for combustion of gasoline. Gasoline is a mixture of many different hydrocarbons, however, the formula can be approximated to C7H16.\n\nThe combustion equation is then:\n\nC7H16  +  11O2 --->  7CO2  +  8H2O\n\nSo, 1 mol of C7H16 or 100 g (7 x 12 + 16 * 1) requires 11 volumes of O2.  Now O2 is only 20% of air, so 5 x as much air is required, ie 55 volumes of air. To work out how much air is need to burn 1 pint of gasoline, you need to know the density of gasoline, which is about 0.73 Now 1 pint = 0.5682 L and at a density of 0.73, the mass of 1 pint of gasoline is 0.73 x 0.5682 = 0.415 kg or 415 g. So volume of air is:\n55/100 * 415 * 24.5 L at 1 atmosphere pressure and 25 deg C or 5,500 L (or about 1200 pints)
784827
pressure(p)=6310.8mm Hg\n             =8.3 atm\nvolume(v)=175 ml\n              =0.175 lit\nr=0.0821\ntemp(t)=295 K\n\npv=nrt   \non solving we get \nno of moles(n)=0.06\n\nnow n=mass/mol wt\nso mol wt=mass/n\n               =6.97/0.06\n               =116.17\n  \n100 gm of compound contains 62.04 gm of C\n\n100/116.17 mole of compound contains 62.04/12 mole of C\n\nso 1 mole of compound contains (62.04/12)x(116.17/100) moles of C ie 6 moles\n\nso 1 molecule of compound contains 6 C atoms \n\nsimilarly 1 molecule of compoumd contains 12 H atoms and 2 O atoms\n\nso molecular formula is C6H12O2\n\nhope that solves your query
165982
Propane - C3H8 = 44 grams/mole\nButane - C4H10 = 58 grams/mole\n1 cc = .001 liter\ngas constant R = .08206\n\nA)\n\nPV=nRT\n(1 atm)(50x.001L) = n(.08206)(273K)\nn=.002 mol\n\n.002 mol propane x (44 g/mol) = .088 g\n.002 mol butane x (58 g/mol) = .116 g\n\n50% propane by mole\n43% propane by mass\n\nB)\n\n(150 kPA)/(101.325 kPA) = 1.48 atm\nPt = P1 + P2\n1.48 atm total = .74 atm propane + .74 atm butane\n\nC)\n\nThe propane should effuse faster because its molecular mass is lower.
913627
Hey, i'm doing that for Chem coursework too! AS right?\nanyway, im not entirely sure, but the most important bit you need is a gas syringe to measure the volume of gas.\nOther bits u will need r:\n-conical flask on tripod, with bunsen underneath.\n-A bung with a glass tube in, to put in the conical flask.\n-This tube leads to the gas syringe, via a rubber seal.\n-Remember the CaCO3 goes in the conical flask.\n-Now all you need to do is heat the CaCO3.\n\nYour welcome!
137179
you need the balanced rxn which has a 2:1 ratio of Mg: O2\n 314 gms Mg = 12.9 moles divided by 2 = 6.46 moles O2. At STP that = 144.7 L volume.
620743
Done with a little bit more precision:\n\nMW of Al2O3 = 101.961 g/mole\n\n% oxygen = 100% * (3*15.9994 g/mole)/101.961 g/mole = 47.08%\n\n% error = 100% * (47.08 - 40.0)/47.08 = 15.03%\n\nKemmy's answer is probably more than sufficient for your homework.  Good job, Kemmy!
1175985
Balance the Carbons first, then the hydrogens, and lastly the oxygens:\nC3H8 + ?O2 --> 3CO2 + 4H2O\n(balances carbon and hydrogen)\nthen:\nC3H8 + 5O2 --> 3CO2 +4H2O (final answer)
836475
1 mole of carbon = 12 g\n(12 g)/(3.51g/cm^3) = 3.4188 cm^3
807508
Most of the info provided is superfluous.\nSimply put, you burned something and got 4.161g H2O (MW=18.016), all H coming from the sample.\nSo you formed 4.161/18.016 = 0.23096 mols H2O, and since each mol H2O contains 2 mols H, your sample contained 0.4619 mols of H.\n\nBTW, it is easy to deduce the empirical formula of the compound:\nNo hydrocarbon can have a MW of 60.095, and since only CO2 and H2O were formed, the compound  must contain O...this works out to be C3H8O, which means it is probably propanol (n or iso) or methyl ethyl ether.
903485
in C6H6 = 6 6 0\n    C14H10 = 14 10 0\n    C8H10 = 8 10 0\n    C6H4(CH3)2 = 8 10 0\n \nand in total C=36\n                   H=36\n                   O=0