How many moles of KClO3 are required to produce 12 moles of O2?
352923
Come on.\n\n2 mol KClO3 = 3 mol O2\nSo 2*4 mol KClO3 = 3*4 mol O2\nSo 8 mol KClO3 = 12 mol O2\n\nAlgebra is fun! lol...not!
832542
Balance the Equation\n4NH3 + 5O2 --> 4NO + 6H2O\n\nConvert from g O2 to moles O2\n\n(64.00 g O2) * (1 mol O2)/(32.00 g) = 2 mol O2\n\nConvert from moles O2 to moles NO\n\n(2 mol O2) * (4 mol NO)/(5 mol O2) = 1.6 mol NO \n\nConvert from moles NO to grams NO\n\nbut first we need to figure out NO's molecular weight\n\n1 mol N * 14.01g/mol = 14.01g\n1 mol O * 16.00g/mol = 16.00g\n                                        30.01g/mol NO\n\nNow we can continue\n\n1.6 mol NO * 30.01g/mol = 48.02 g NO produced\n\nHope this helps!
699263
V=nRT/P\nn(O2)=3/2(0.0022mol of KClO3) =0.0033 gmol\nR=0.0826\nT=294 k\np = p (O2) + p (water vapour)\np(O2) = 0.926 Atm\nV=(0.0033)(0.0826)(294)/0.926=0.087 liter = 87 ml
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
761151
first you must calculate the number of moles for oxygen, with its molar mass of 32.00g/mol.\n\nonce you have that value, you know from the balanced equation that the number of moles of SO3 will be equivalent to twice the number of moles of oxygen.\nusing this value for moles of SO3, you can then find the mass, knowing that SO3's molar mass is equivalent to 80.07g\nthe mass of SO3 produced will then be 0.9280*mass of SO3. (92.8% yield means only 92.8% of expected was produced)
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!
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.
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
119509
my question is:\n\nNaHCO3(s) + NaOH(aq) -> Na2CO3(aq) + H2O(l)\n\n3.25 moles of Na2CO3 is produces.
619470
o.k., in short, d < a < c < b. This is basic material, so I'll give a generously detailed explanation.\n\na) 2 moles N2 * (28 g N2/ mole N2) = 56g N2\n\nb) 1 mole KCl * [(39.1gK +35.5gCl)/ mole KCl] = 74.6 g KCl\n\nc) 9 x 10^23 molecules CO2 * (1 mole CO2/ 6.02 x 10^23)*([12gC + 16gO + 16gO]/ mole CO2) = 66g CO2\n\nd) 2 moles LiF * [(6.94g Li + 19gF)/ mole LiF] = 51.88 g LiF
768474
3.24 kmol= 3.24*10^3mol=3.24*10^3*261.992 [molecular wt of the compd.] g=848854.08 g
878166
To answer this, you can use the ideal gas law, V = nRT/P\n\nYou know R , T  and P, andso you need to find n.\n\nTo find n , note that 2 moles of KClO3 produces 3 moles of O2. Given the mass of KClO3, you kind find the moles of this compound by dividing the mass by the molar mass. Assuming that there is 100% conversion, you can use the ration above to find the moles of O2.\n\nLastly, I'm not sure, but I think you need to use the gage pressure, which is the barometric pressure - the vapor pressure.
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.
878226
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.
750464
My chemistry is pretty rusty, but let's see.\n\nI'll assume that you have an infinite amt of O2, and that you have 68.12 grams of NH3.\n\nThe equation isn't balanced, so you need to do that first (you have 3 hydrogens on one side, but two on the other).  Balance to get:\n\n2(NH3) + (2.5)(O2) --> 2NO + 3(H20)\n\n4(NH3) + 5(O2) --> 4NO + 6(H20)\n\nNow, given that you have 68.12 grams of NH3, you need to convert this into moles of NH3, as Stoichiometry deals with molar ratios.  To get the conversion ratio, look at the periodic table.  Under each element, there should be an atomic mass.  For oxygen is 16.something, Hydrogen 1.008 ( i think ).  Find it for N and H.  (this is where I get hazy on the math).  I believe: take 3 * atomic mass of hydrogen + atomic mass of nitrogen, let's call it M  (it should be in grams/mole)  So to get moles, multiply 68.12 * (1/M).  \n\nNow, our equation above shows that for every 4 moles of NH3, we get 6 of H20.  # moles NH3 * (6/4) = # moles of H20.  Now, do the reverse operation of H20 to get how many grams is produced.  H20 has an atomic mass of something like 18.016.  So, # moles H20 * 18.016 = # grams H20\n\n\nI hope that's somewhat correct, and if not, leads you in the right direction at least
844713
I don't know how many significant figures you need them in, but here's how you do it. You find the molecular weight of MnO2. You find that it is 86.9 grams. So you divide the 5 grams you have by that molecular weight to give you number of moles (you should get about .0575374 moles). Now you take that, and using the mole ratio of 4 moles HCl / 1 mole MnO2 (essentially multiplying by 4), you get about.23015 moles HCl. Now you take that number, and multiply it by the molecular weight of HCl, which is about 36.5. \n\nYou should get about 8.4 grams. Hope this helps, and keep up the good work.
945679
When  excess of SO2 is present the amount of product will only be controlled by amount of Cl2 present.\nNow, 2 moles of  Cl2 forms 1 mole of Cl2O\nThereofre, 1 mole of Cl2 will form (1/2) i.e. 0.5 moles of Cl2O\nThus, 0.400 moles of Cl2 will form 0.200 moles of Cl2O.\n\nHence answer is C.