How many moles of water can be made from 4 moles of oxygen gas and 16 moles of hydrogen gas? What is the limiting reagent?\n\nI'm basically just looking for the answer to check against mine and if it's wrong then I'll know what to change in the problem. I'm just having some confusion over the setup of this problem. Thanks.
217085
you need 2 moles of hydrogen for each mole of oxygen so for four oxygen moles you need eight hydrogens. oxygen is the limiting reactant and you can make four water moles
309425
Okay, this should be easy:\n\nHow many moles of each substance do you have?\n\n70.25 g of steam (water vapor) = (70.25 g)/(18.0 g/mole) = 3.90 moles\n\n1.470 g of H2 = (1.470 g)/(2 g/mole) = 0.735 moles\n\n6.58 g N2 = (6.58 g)/(28 g/mole) = 0.235 moles\n\nTotal number of moles you have is = 3.90 + 0.735 + 0.235 = 4.87 moles\n\nSo, the mole fraction of steam is 3.90/4.87 = 0.801 or 80.1%\n\nHydrogen: 0.735/4.87 = 0.151 or 15.1%\n\nNitrogen: 0.235/4.87 = 0.048 or 4.8%
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.
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.
160066
For example, hydrogen burns in oxygen to form water vapor. \n\n2H2(g) + O2(g) -> H2O(g) \nHow much hydrogen must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 3000C? \nTo work this problem, we need to figure out how many moles of water vapor we need to form. Use the ideal gas equation to compute this: \n\nPV = nRT \n1.00 atm * 1.00 L = n*0.0821 (L*atm/mol*K)* 473 K \nn = 0.0258 moles \nThus, we need to form 0.0258 moles of water. Now it's just a basic stoichiometry problem: for each mole of water formed, you need 2 moles of hydrogen gas, so 0.0258 moles water * 2 moles H2/1 mole water = 0.0516 moles of hydrogen \n\n=====================================================\n\nthis will help you a lot\n\nhttp://www.chem.hawaii.edu/asato161/Ch10NotesP2.pdf#search='Gas%20Stoichiometry%20PV%3DnRT'\n\nhttp://members.aol.com/profchm/gasstoic.html
130624
I'm not exactly sure what you're talking about, but yes, water can be separated into hydrogen gas (H2) and oxygen gas(O2). It's the following redox reaction.\n\nreduction: 2e- + 2H+ + H2O --> H2 + H2O\noxidation: 2H2O --> O2 + 4H+ + 4e-\n\nMultiply the reduction reaction by two and add the two together to get the final reaction.\n\n2H2O --> 2H2 + O2\n\nThe hydrogen goes down from a +1 state to a 0 state, and the oxygen goes up from a -2 state to a 0 state.
427393
The equation is  2H2 + CO--->CH3OH\nThe first step in a problem like this is to determine if you have enough of each reactant...or is there a limiting reactant.\n\nYou started with...\n1.73g H2 (MW 2.016) = 0.858mols...and\n16.05g CO (MW28.01) =0.573mols\nBut since 0.848mol H2 would react with only 0.429mols CO you have an excess of CO...too much CO, so your limiting reactant is H2.\nYou get one mol of CH3OH (MW 32.04) for each half mol of H2 per equation, so the theoretical yield is...\n0.429 x 32.04 = 13.7g of CH3OH\n\nCheck my figures for accuracy...which will also give you good practice. :-))\n\nEDITED LATER: If you check this work carefully, you will note that the answer is correct, but the statement is not...it should read:\nYou get one-half mol of CH3OH for each mol of H2 per the equation, so the theoretical yield is...\n1/2 x 32.04 x 0.858 = 13.7g CH3OH\n(Did you find this error?)
542832
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 6.406 g x (12 / 44) = 1.747 g\nhydrogen: 5.244 g x (2 / 18) = 0.583 g\noxygen: 4.664 - 1.747 - .583 = 2.334 g\n\n2) convert to moles.\ncarbon: 1.747 / 12 = 0.146 mol\nhydrogen: 0.583 / 1 = 0.583 mol\noxygen: 2.334 / 16 = 0.146 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.146 / 0.146 = 1\nhydrogen: 0.583 / 0.146 = 4\noxygen: 0.146 / 0.146 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is CH4O.\nThe atomic masses add up to 32, which is also the molar mass of the compound.
531824
First you have to know the chemical equation, which is 3H2 (g) + N2 (g) <=> 2NH3 (g)\n\n\nThen, you calculate the number of moles for each reactant that is available, using the mass given.\nno of moles of H2\n= mass of H2 / relative molecular mass of H2\n= 2.55g / 2 x1                       ---> there are 2 H atoms in H2\n= 1.275\n\nno of moles of N2\n= mass of N2 / relative molecular mass of N2\n= 13.65 / 2 x 14\n= 0.4875\n\n\nThen you need to find which is the limiting reactant. In this case, the limiting reactant is H2. The reaction will stop due to the lack of H2 to react with N2, even though there is excess N2.\n\nUsing the no of moles of H2,\nno of moles of NH3\n= (1.275 / 3) x 2               ----> make use of the mole ratio\n= 0.85\n\nThus, calculating the mass of NH3,\nmass of NH3\n= no of moles x relative molecular mass\n= 0.85 x (14 + 3 x 1)\n= 0.85 x 17\n= 14.45g\n\nBy the way, it is called theoretical yield because it is highly impossible that the amount you collect at the end of the experiment is the exact amount you have calculated. :D\n\nCheerios.
448333
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.847 g x (12 / 44) = 2.686 g\nhydrogen: 5.038 g x (2 / 18) = 0.560 g\noxygen: 4.146 - 2.686 - .560 = 0.900 g\n\n2) convert to moles.\ncarbon: 2.686 / 12 = 0.224 mol\nhydrogen: 0.560 / 1 = 0.560 mol\noxygen: 0.900 / 16 = 0.056 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.224 / 0.056 = 4\nhydrogen: 0.560 / 0.056 = 10\noxygen: 0.056 / 0.056 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
537979
1) Determine how many moles of CH4 we have:\n175 g/ (12.01+4*1.008) =10.9 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nCH4 + 2 O2 ->  CO2 + 2 H2O\nWe need 2*10.9 = 21.8 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. At the given conditions:\nV(1mol) = 22.4 * (290/273) = 23.79 liter\n\nTherefore 21.8 mol = 519 liter = 0.519 m^3
243020
here, you need to apply the concept of limiting reagent.\n\nhere, N2 is limiting reagent. \n\nSo, if 28 gm (molar mass of N2)nitrogen would give you 34 gm of (molar masss of 2NH3) ammonia.\n\nSo, 3 gm (given) nitrogen would give you how many?\n\n = (7 * 34)/ 28\n\n = 8.5 gm of ammonia.
421940
Since it takes 1 mole of O2 to react with H4 and get H2O + H2O. I would have to say that .264 moles of H2O were created. water is 18gper mole so .264*18g=4.752g of water.
514121
find out how many moles of each gas...\n\nfind the mole ratio....\n\nsince mole ration is directly proportional to pressure ratio...\n\nyou can find out partial pressure of H2...\n\n5.97 atm....\n\n\n?
536609
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
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.
424618
Cool stuff, I used to get confused by ugly chemistry calculations, too :D\n\nYou have 3.88 g acid in 250cm3, so in the 25 cm3 you titrated there are 0.388 g of the acid. You need to know how many moles that is, so that you can calculate molar mass.\n\nThe amount of NaOH used up is 46.5 cm3 × 0.095 mol/dm3, that is 0.0465 dm3 × 0.095 M, which equals 0.00442 mol (It's not exactly that much, but I'll stick with three significant figures).\n\nSince NaOH reacts with monoprotic acids in a 1:1 ratio, that means that there was exactly the same amount of acid in the solution.\n\nNow we know that 0.388 g of the acid is 0.00442 mol, so we can divide the mass by the number of moles to get molar mass. That is 0.388g/0.00442 mol = 87.8 g/mol.\n\nRelative molecular mass is simply molar mass in grams, but without the unit, so Mr of the acid is 87.8.