What volume of O2 (m3) is needed for the complete combustion of 117 g of   C3H8 at 14.0oC and 100.0 kPa?
930613
CHEMICAL EQUATION\n\nC3H8 + 5O2 --> 3CO2 + 4H2O\n\n117g C3H8 * (1mol C3H8 / 44.09652g C3H8) = 2.65mol C3H8\n2.65mol C3H8 * (5 mol O2 / 1 mol C3H8) = 13.3 mol O2\n\n14.0°C + 273.15 = 287.15K\n\n 100.0 kPa (1 atm / 101.325 kPa) = .9896 atm\n\nPV = nRT\nV=nRT/P\n\nV= (13.3 mol*.0821L atm mol-1 K-1 * 287.15K) / (.9896 atm) = 317 L O2
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.
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!
970312
If you can figure out how to calculate the number of grams of O2 it takes, then you just need to us Pv = nRT to get the volume. I forget the equation for combustion though...need a combustion/engine textbook.
942373
From the reaction\n\nS8(g) ===> 4S2(g)\n\nKp = (pS2(g))4/pS8(g) where p is the partial pressure\n\nGiven the partial pressure of S8(g) at equilbrium 0.25 atm, clearly 0.75 atm of S8(g) has been consumed to produce (from the equation) 4 volumes of S2(g), Given the volume was fixed this mean dS2(g) = 3 atm\n\nSubstituting into Kp yields\n\nKp =3^4/0.25 = 324\n\nThus Kp at 1325 K is 324
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.)
866626
O3 = 16g x 3 = 48g/mol.\n 1 Mole O3 = 48g  (÷ 4)\n 0∙02083333...Mole = 1g    (x 0∙740)\n 0∙014516666.Mole = 0∙740g\n\nNO = 14g + 16g = 30g/Mol.\n 1 Mole NO = 30g\n 0∙03333...Mole = 1g\n 0∙022333.. Mole = 0∙670g\n\n If the O3 splits to give O2 + O,\nthen the reaction is 1:1, and O3 is the limiting reactant.\n\nExcess NO: \n = 0∙022333.. Mole of NO  -  0∙014516666.Mole of O3\n = 0∙0068666...Moles of NO.\n\n If the O3 splits to give O2 + O, and splits again to give O + O + O, then the reaction is 3:1, and NO is the limiting reactant.\n\nExcess O3:\n = 2/3 x  0∙014516666.Mole \n = 0∙009677777....Moles of O3.
936398
To understand this question, you need to know what kind of product will be produced during a combustion. Generally, for a complete combustion of a hydrocarbon (contains only H and C), the product will be H2O and CO2. Depending on the C and H content in the hydrocarbon, you will produce different amount of H2O and CO2. \n\nOn the other hand, if it is an incomplete combustion (with limited supply of O2), apart from CO2 and H2O, you may also produce C and CO. The relative amount between them varies depending on the amount of O2 that is made available. With very limited amount of O2, more C (soot) will be produced. This leads to the "sooty" flame you are seeing.\n\nIn order to explain why generally alkenes require less O2 to burn compared to alkanes, let's compare ethane (2 C alkane) and ethene (2 C alkene). \n\nEthane has a formula C2H6 while ethene has a formula C2H4. Their corresponding balanced equation for combustion are:\n\nC2H6 + 7/2 O2 → 2 CO2 + 3 H2O\nC2H4 + 3 O2 → 2CO2 + 2 H2O\n\nAs you see, since both compounds have same number of C, they both produce 2 mole of CO2 for every mole of compound. However, since they have different number of H, the number of moles of H2O produced will be different. Since H2O has a fixed formula, in order to produce it, you need to supply O2. That's why the two compounds require different amount of O2.\n\nThe reason that you see a sooty flame is due to the fact that you don't have enough oxygen supply during the burning process. You can easily confirm this by using just a Bunsen burner (but be very carefully!!!). First, turn on your Bunsen burner with the air hole closed. The flame you see at the moment should be a luminous flame. It is golden yellow color. Now, if you carefully put a glass plate (hold with clamp or other device) on top of the flame and slowly move towards the tip of the burner (but don't cover it completely), you should see the plate will get "darker" by the flame. This is the soot produced during the combustion. Now, take away the plate, open the air hole in the burner to the max. You should see a the flame turns to blue. If you put the glass plate (with the soot marks) back to the flame, you can see the flame can slowly "clean up" the plate (if enough O2 is supplied with the flame) and the soot will disappear. The soot is turn back to CO2 due to the excess of O2.
702782
CH2OHCl: 66,45g/mol\nDo you mean 4dm^3 or a cube with 4dm length  edges?\nIf you mean 4dm^3 then\n4dm^3= 4*10^3 cm^3 = 4*10^3 mL= 4L\n2M= ?/4L\n?= 8 moles of CH2OHCl \nif you meant 4dm edges then\n4dm= 4*10^1cm\n(4*10^1cm)^3= 6*10^4cm^3 (actually it is 6,4 but because of the lack of figures you must report 6 instead of 6,4)\n6*10^4cm^3= 6*10^4 mL = 6*10^1 L\nTherefore\n2M= ?/6*10^1L\n?= 1,2*10^2 moles of CH2OHCl \n\nPart II\nCH4 + 2O2 → CO2 + 2H2O\n7dm of O (again: 7dm edges or 7dm^3???)\nfor 7dm sides\n(7*10^1cm)^3= 3*10^2 L \nO density = 1,43 g/L\nTherefore: 5*10^2g or 0,5 kg of Oxygen\n4g CH4*(1mol CH4/16g CH4)*(1mol CO2/1mol CH4)= 1*10^1g of CO2\nIf you compare the amount of Oxygen and methane and the molar ratios there is no need to do the same for O. Obviously CH4 is the limiting reagent\nCO2 density: 1,98*10^-3 g/mL\n1,98kg/m^3\nIf 5dm (again the same problem) were produced, then\nfor 5dm edges: 2*10^2g of CO2 (obviously not this one)\nfor 5dm^3: 5*10^3 cm^3 of CO2: 1*10^1g of CO2 this is it!)\nNow the yield\nIf you would have added more figures we would have had\n(9,9g/11g)*100= 90% yield\nBut because of the lack of figures we end up with\n(1*10^1g /1*10^1g)*100= 1*10^2 % (you got the maximum yield)
141442
KE per mole = (3/2) RT where R = 8.3145 J/mole*K and T is temperature in kalvin, so:\n\nKE = (3/2) RTn where n is the number of moles of gas\n\nYou can solve PV=nRT for n and substitute to get KE = (3/2) PV where P is pressure and V is volume.
818293
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
752459
This is a complete combustion.\nReaction:            aX + bO2 --> dCO2 + eH2O\nd = 0.811818182, derived from mass of CO2 and its Mr.\n\nX has 0.811818182 mol of C.\n\nOriginal sample has 0.811818182/ 1.626 x 60.095\n                                  = 30.00382143mol of C
602334
PV = nRT\n\nV = 3 L\nn = g CO2/MW CO2 + g O2/MW O2\nT = 273 K\nR = 0.082057 L-atm / mol-K\n\nsolve for P
963470
CO + 2H2  --->  CH3OH\n\nmoles H2: 2.00g/ 1g/mol = 2.00 mol\nmoles CO: 18.50g / 28g/mol = 0.6607mol\n\nyour limiting reactant is CO since for every 0.6607 moles of CO, you'll only need 1.321 moles H2. there's an excess of 0.6790 moles H2.\n\ntheoretical yield, therefore, is (1.321/2)*100 =  66.05%\n\nbased from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,\n\nweight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams
739541
CH4 + 2O2-> CO2 + 2H2O
770748
Oxygen is the limiting reactant....you would need 5.6 moles of it for a complete reaction.\n\nMoles of CO2...2.8.  It is 1/1 with methane\n\nGrams 2.8 X 44 = 123.2 grams
832153
2 moles of reactant  (C2H4F2) give you 4 moles of HF\nC2H4F2 : 66 g/mol\nHF: 20g/mol\nTherefore\n79,3g  C2H4F2*(1mol  C2H4F2/66g  C2H4F2)*(4 moles HF/ 2moles C2H4F2)*(20g HF/ 1mol HF)= 48,1g HF\nThe amount of HF you are expected to get with 100% yield is 48,1g. If you got 18g, then\n(18g/48,1g)*100= 37% yield