How many moles of water, H2O, can be produced from the reaction of 16 grams of methane, CH4, with an unlimited supply of oxygen, O2?  How many grams of water is this?  The reaction is:\n\nCH4 + 2 O2 Þ CO2 + 2 H2O
212047
No, this is easy and fredy got it right.\n1 mole of methane yields 2 moles of water.\n2 moles of water weighs 36 grams.
528727
This is easy, previous answer had a mistake, so I decided to write.\n\nYou can use the equation for gas state\n\nPV = (m/M)RT, where m = the mass you are asked to determine, M = molar mass of water molecules, \nM = M(o) + 2M(h) = 18 g/mole,\nP is pressure, we shall trasform torr to Pa, which is Pa=N/sq.m.\nP = 4.6 torr = 4.6x133.322 Pa = 613.3 Pa\nV is volume, \nV = 0.75 L = 0.00075 cubic meters\nT is temperature,\nT = 0 C = 273 degrees K\nR is a Universal Gas Constant\nR = 8.314 J/(mole x deg.K)\n So,\n m = MPV/(RT) , all units are in SI system, M and m in grams or kilograms simultaneously, and we can find m:\n\nm = 18 (g/mole)x613.3(Pa)x0.00075(cub.m.)/(273(deg.K)x8.314(J/(mole x deg.K))) = 8.28(gNm/mole)/2269.7(J/mole) = 0.00365g , here we have noticed that Nm (force in N multiplied by distance in m) gives energy in J, and the final dimensions give grams\nSo, m = 0.00365 g = 3.65 miligrams\n\nThat is the correct answer, you can trust
554721
You have 2 38.0 mL volumes, so the total volume is 76.0 mL.\nThe reaction is H2SO4 + 2 KOH --> 2 H2O + K2SO4\nThe rise in temperature (delta T) is 30.17°C - 23.50°C = 6.67°C\n\nThe heat capacity of water is 1.0 cal/(g*°C)\nThe density of water is 1.0 g/mL\n\nThe H of the reaction is 1.0 cal/(g*°C) * 1.0 g/mL * 76.0 mL * 6.67°C = 506.9 cal or (506.9 cal * 4.186 J/cal * 0.001 kJ/J) = 2.122 kJ\n\nThere were 0.82 mol/L of KOH, and 38.0 mL, so there were .82*.038 mol = 0.03116 mol of KOH\n\nDeltaH per mole of KOH is 2.122 kJ / 0.03116 mol = 68.1 kJ/mol
345087
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)
441998
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 4.203 g x (12 / 44) = 1.146 g\nhydrogen: 3.441 g x (2 / 18) = 0.382 g\noxygen: 3.060 - 1.569 - .262 = 1.532 g\n\n2) convert to moles.\ncarbon: 1.146 / 12 = 0.0955 mol\nhydrogen: 0.382 / 1 = 0.382 mol\noxygen: 1.532 / 16 = 0.0958 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.0955 / 0.0955 = 1\nhydrogen: 0.382 / 0.0955  = 4\noxygen: 0.0957 / 0.0955 = 1\n\nThe empirical formula of the substance is (CH4O)n\nThe atomic masses add up to 32, which is also the molar mass of the compound. Therefore n=1 and we find CH4O.
111731
ur equation is wrong....\nu will get H2O + C...\ncheck again\nif it's correct, u have to find out which one is the limiting reagent by calculating the number of moles of each reactant used...thus figuring out the number of moles of product and multiplying with the molecular mass of methanol...
397393
Reactants: Hydrogen and Oxygen gas are used up\nProducts: water is produced\n\nH2(g) + O2(g) -----> H2O (l)\n\nBalance O: H2(g) + O2(g) -----> 2H2O (l)\n\nRebalance H: 2H2(g) + O2(g) -----> 2H2O (l)
132882
Extracting Methane from Water?\n\nUsually, there isnt much methane in water.  Methane, being a non-polar Hydrocarbon, does not readily dissolve in polar solvents like water.\nHowever, if by chance you actually did have some CH4 dissolved in water, a gas' solubility in water decreases as pressure decrease and temperature increases, so by boiling the water and/or lowering the atmospheric pressure above the liquid, the methane can be removed.
77932
To solve this problem we need to think about limiting reactants.  This one is fairly simple because I already have moles and the ratio is 1:1:1.  If I only have 3 kmols of C, that is the most I can make, because going beyond that, I would be out of C.  Thus, I can only make 3kmols of CO2 with 2 kmols of O2 left unreacted.  These amounts are theoretically find for a combustion reaction.
474542
The method is very simple.  I can do the first and you can try the others.\n\n1.  First write the reaction and balance it.\n\n2HgO ---> 2Hg + O2.\n\n2. Convert grams to moles of whatever mass you have.  In this case you have O2 as 96 grams.  You convert to moles by dividing the mass by the molecular weight.  In this case, for O2 will be 32 g/mol\n\n96 grams / 32 g/mol = 3 mol.\n\n3. Now you look back to the equation.  In the equation for producing 1 mol O2 you need 2 moles of HgO.  If now have 3 moles O2, then you will need 2 x 3 moles of HgO, 6 moles.\n\n4. Now you need to convert the moles of HgO to grams.  To do this calculate the molecular weight of HgO.  It is 216.6 g/mol\n\n6 moles * 216.6 g/mol = 1299.6 g of HgO.
339420
The first thing you need is a balanced equation for the reaction:\n\n4 NH3 + 5 O2 → 4 NO + 6 H2O\n\nA) 5 moles of O2 produce 4 moles of NO\n\n160.00 g of O2 / g/mol for O2 = number of moles of O2\n\n160.00 g / 16.00 g/mol = 10.0  moles of O2\n\nThis would produce 8.0 moles of NO\n\n8.0 moles * 30.01 g/mol = 240.08 grams of Nitric oxide.\n\nB)  If the percent yield is only 92.98 % then\n\n240.08 g * 0.9298 = 223.226 grams (you might want to round this to 223.2 grams).
419467
ooo this will be fun!\n\nok..\nA)\n2.207g of H2O MW. 18.0\nmol = weight of sample (g) / molar weight (g/mol) (n=m/M)\nn = 2.207 / 18\nn = 0.1226 moles of H2O\n\nnow, since there is 2 hydrogens in H2O, then when the molecule is broken down, there is twice as many moles of hydrogen then the overall molecule,\ntherefore,\n\n2x 0.1226 = 0.2452 moles of H.\n\ndetermining number of moles of CO2 is also helpful,\n4.313g, n = 4.313 / (12.01 + 16 +16) = 0.0980moles\napprox, 5(moles H2O) :  4(moles CO2)\n\nB) \nworking the number of moles of this compound will be useful:\nn = 1.816 / 74.122 = 0.0245 moles, which is 1/10 of the moles of H produced. (or 1:5 of the moles produced of water)\n\nif you think about it, the reaction is something like this:\nCH2 + O2 --> CO2 + H2O\nclearly the above doesn't make sense, because 1) it isn't balanced, and 2) the reactant needs a higher molecular wt.. now thinking about it, the following can be deteremined:\nand from the ratio of moles found above:\n       1     : (excess):  4        :    5\nC4H10 + 13/2O2 --> 4CO2 + 5H2O\n\nnow, I amoung most allow fractions in equations, but if you don't, multiply through by 2.\n2C4H10 + 13O2 --> 8CO2 + 10H2O\n\nnow, you want the empirical forumla,\ntherefore,\ndivide by highest common factor,\nC2H5 \n\nthe empircal fomula is just a ratio and doesn't make huge sense, as the formula for an alkane is CnH2n+2\nhope that makes sense :)
465943
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.
402410
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.
313427
If oxygen is your limiting reactant, then the amount of CO2 produced is dependent solely on the amount of oxygen.\n\nThe balanced equation being 2CO + O2 ---> 2CO2, you will generate 2 moles of carbon dioxide for every mole of oxygen, so the answer is 100 moles.
75522
The thing with gasses is that although one mole of gas is still the atomic weight in grams (watch for the diatomic gasses), with a gas at STP that amount amount of gas always occupies 22.4 liters.\n\nSo you have a new conversion factor:  1 mole = 22.4 liters.  Use it just like the molar mass (GMW? - whatever it's called these days).\n\nE.g. for A, 40.86 L of any gas - methane included - is \n48.89/22.4 moles of the gas.  From there on it's just like problems you've worked before.  From the equation, one mole of methane requires 2 moles of C, so about (48.89/22.4)(2)(12.01) grams.
29204
147.8 g of CO remains.\n\nFor every 28 g of CO you need 4 g of H.\n\nSo if 12.1 g of H2 reacts then 84.7 g of CO are consumed. That leaves 147.8 g of CO unreacted.