how many grams of   H2O   are produced from the complete reaction of   64.16 g   of   CH4 ?
961596
You first balance this equation:\n\nCH4 + 2O2 ---> CO2 + 2H2O\n\nTotal number of mole of 64.16 g CH4:\n= 64.16 / (12 + 4)\n= 4.01 moles\n\nTotal number of moles of H2O produced:\n= 4.01 * 2 = 8.02 moles\n\nTotal mass of H2O:\n= 8.02 * (2 + 16)\n= 144.36 g
446794
1 kg of water = 1000 grams of water.\nWater (H2O) has a molar mass of 18.02 grams per mol.\n1000 grams of H2O equals about 55.5 moles of water.\n\nThe reaction which water is split into Hydrogen gas and Oxygen as is as follows,\n2H2O (l) --electrolysis--> 2H2 (g) + O2 (g)\n\nThe change in the enthalpy of formation of this reaction equals about +571.8 kJ, meaning that at standard conditions and at a constant pressure, 571.8 kJ of work must be done on the system to totally split 2 moles of water into 2 moles of H2 and 1 mole of O2 gas.\nUsing this information, we find that to split 55.5 moles of water into 55.5 moles of H2 gas and 27.75 moles of O2 gas, it would require 15867.45 kJ of energy (or about 1.6 E7 Joules).\n\nThe amount of energy needed to split the water is also the maximum amount of energy which will be obtained if the H2 and O2 are then allowed to recombine again into water.
781803
--->H2O  +  H+  +  Cl-
949367
First you have to get the molecular weight of those.\n\nits\n180 for C6H12O6\n192 for 6O2\n264 for 6CO2\n108 for H2O\n\na. Since you have the same amount of C6H12O6 and O2, your limiting reagent in this case is the O2.\n\nSo you use it to determine the CO2 produced.\n\nb. if 192 grs of O2 produce 264 grs of CO2\nthen 100 grs will do: (100 * 264)/180 = 137.5 grs of CO2.\n\nc. Now, to calculate the percent yield, in ideal conditions you would have gotten 137.5 grs (which is your 100% theorically) but practically you got 97.1 grs. so\n\n137.5 ___ 100%\n97.1______ ?\n\n(97.1*100)/137.5 = 70.62% is the perecnt yield of CO2.\n\nd. You do the same thing backwards to determine how many excess of the non-limiting reagent (C6H12O6) you have.\n\nto produce 264 grs of CO2 you need 180 grs of C6H12O6\nfor 137.5...\n\n(137.5*180)/264 = 93.75 grs of C6H12O6\n\nso the excess is 100 - 93.75 = 6.25 grs of C6H12O6
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
518927
(46.2 grams)/(32 grams per mole) = 1.44 moles of Oxygen (O2)
270274
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...
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
920990
methane: CH4\n\nMr;\ncarbon 12\nhydrogen 1\n\nMr of methane= 1*carbon+4*hydogen\n=12+4\n=16\n\ntherefore one mole of methane weighs 16 grams\n\n4 g of methane= 4/16\n=0.25 moles\n------------------------------...\n\none mole of any gas occupies 24 dm3 at r.t.p\n\nmoles of oxygen in 7 dm3= 7/24\n=0.292 moles
112430
193,000 Amperes per second to produces 1 mole of H2 or 2.016 grams of x.
984937
The equation is: 2H2 + O2 ---> 2H2O\nSo you would form 2L of water VAPOR\nIf you want this in terms of LIQUID water, remember that one mol of a gas occupies 22.4L, so you have 2/22.4 mols of water (MW=18) formed, so...\n2/22.4 x 18 = 1.6 grams or 0.0016L of LIQUID water.\nThis assumes that all three gasses act as 'ideal gasses' (which is only close) and conditions are reduced to STP...so answer is not exactly correct, but 'close enough for government work'  :-))
995159
Use the atomic weights of the individual elements, and add them together.  For instance, if I had water, H2O:\n\nH - 1.01 g * 2 = 2.02 g\nO - 16.00 g\n\nSo the molar mass of H2O would be 18.02 g/mol.\n\nIf I wanted to convert from grams to moles, I would divide the grams by the molar mass.  Grams cancels out to be left with moles.\n\nIf I wanted to convert from moles to grams, I would multiply the moles times the molar mass.  Moles would then cancel out to be left with grams.
707517
The 4g of methane is a quarter of a mole. ( 1 mole of methane is 16g). A mole of ANY gas occupies 22.4 litres at normal temperature and pressure. So a quarter of a mole has a volume of 5.6 litres.\nThe equation for the reaction is:\nCH4 +  2O2  =  CO2 +  2H2O\nThis shows that twice as much oxygen is needed as methane.\nSo to burn 5.6 litres of methane you need 11.2 litres of oxygen.\nThis means that the limiting reagent is oxygen as there isn't enough.\nIf 5dm3 of CO2 is made then more oxygen must have been added as you would only expect to get the half the volume of CO2 than the volume of oxygen you started with. In your case you started with 7 dm3 of oxygen and so the maximum amount of CO2 is 3.5 litres.\nAssuming the 4g of methane were burned in EXCESS oxygen you would expect to get 5.6 litres of CO2 and in this case the % yeild is  5 divided by 5.6 multiplied by 100 = 89.3%\n\nAt room temperature the figures would be slightly different. i.e 6 dm3 of CO2 would be expected and the yeild would be 83.33%\n\nSo you can see that it is important to say at what temperatures the gases were measured as they tend to expand when they are warmed.
961444
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.
900861
As usual, we first check for a balanced equation.  It isn't, so we fix it: 2 LiOH + CO2 -> Li2CO3 + H2O.  Also as usual, get the molecular weights: LiOH = 24, CO2 = 44, and the others don't matter.  There are 4 moles of CO2, so we need 8 moles of LiOH, and that is 192 grams.  Extra credit: find the error in the next response.
693257
From your reaction equation, 3 moles of hydrogen gas produces 2 moles of ammonia if the reaction was 100% efficient.  From the way you phrased your question, it's difficult to tell if you mean 450g of hydrogen (H), or 450 grams of hydrogen gas (H2).  I'll assume you mean hydrogen gas.\n\nFirst you have to calculate the number of moles hydrogen gas you started with:\n\nMoles H2 = grams H2 / Molecular weight H2\nMoles H2= 450 g H2 / 2.016 g per mole H2\nMoles H2 =223.21 moles\n\nIf the reaction was 100% efficient:\nMoles NH3 = 2/3 * moles H2\nMoles NH3 = 2/3 * 223.21\nMoles NH3 = 148.80 moles \n\nNow to calculate the grams NH3 expected from 100% efficiency:\ngrams NH3 = moles NH3 at 100% efficiency * mol wt NH3\ngrams NH3 = 148.80 moles * 17.024 grams/mole\ngrams NH3 = 2533.17 grams\n\nSo to calculate the percent yield for the reaction:\n\n% yield = (actual yield/ theoretical yield) *100\n% yield = (1575 grams/ 2533.17 grams) *100\n% yield = 62.17%\n\nNow if you meant H instead of H2 for the original grams hydrogen in your question, you'll have to recalculate as appropriate.
1097704
I believe it's a straight forward mole ratio conversion using the balanced equation.  For every one mole of Mg, one mole of H2 is produced.  \n\n.00164mol H2