what is the total mass of products formed when 16 grams of CH4 is burned with excess oxygen?
930204
First, convert 16 g CH4 into moles.  Given the molecular weight of methane (CH4) is 16, we have one mole of methane on our hands.\n\nNow, find the equation for the reaction.  This is:\n\nCH4 + O2 -->  CO2 + 2 H2O\n\nIf we assume complete combustion of the CH4 (given excess oxygen), then we have one mole of carbon dioxide and 2 moles of water on our hands.\n\nFinally, convert that back to weight using the molecular weight of carbon dioxide (44) and water (18) and you get\n\n44 + 2 (18) = 80 grams of total products.
816363
4 Fe(s) + 3 O2(g) ----> 2 Fe2O3(s)\n\nFirst, you have to check if the chemical equation is balanced. In this case, it is. So, what it tells you is that for every 4 mole of iron that is reacted, 1652 kJ will be released to the surrounding. Now, since O2 is in excess, iron will then be your limiting reagent. If you have 3.34 mole of iron, you can use ratio to determine how much energy will be released.\n\nE = (-1652/4) *3.34 = -1379.42 kJ\n\nNote that the "-ve" sign indicates energy is released instead of being absorbed.
963278
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
552992
ch4 for both.\nOne carbon atom and 4 hydrogen atoms. Simple!
839857
1) The reaction equation is  CO + 2H2 -> CH3OH.\n\n2) Calculate the number of moles available:\nCO: 20.05 g = 20.05 / 28.01 = 0.7158 mol\nH2: 2.02 g = 2.02/1.008 = 2.004 mol\nWe are limited by the amount of CO available and therefore \n0.7158 mol of CH3OH will be produced.\n\n3) Calculate the mass of CH3OH:\n0.7158 mol = 0.7158 * (12.01 + 4*1.008 +16) = 22.93 gram
852876
Assuming STP, one mol of oxygen (64 grams) would be 22.4 L.\n\nSo divide 1 by 22.4 and that's how many mols you have. Multiply by 64 grams and there's your mass of one Liter. It'd be useful to convert to kg now, just multiply by 10^(-3)\n\nWeight is mass multiplied by the acceleration of gravity. Assuming this is on earth, acceleration is 9.8.\n\nSo final answer should be 28*10^(-3)  kgm/s^2==>.028 Newtons\n\nIf you want pounds, multiply by 0.2248 to get 0.0062944\n\nShould be right, although I'm not sure. I'm mixing physics and chemistry...unless you wanted mass, not weight, then ignore everything after mass.\n\nEDIT: Just wanted to point out a mistake in the answer above...he forgot that oxygen gas is O2. Weight of one mol of O is 32 g, but molecular weight of one mol of O2 is 64 g.\n\n2nd EDIT: Just realized...I was wrong, Maussy right. Atomic weight of oxygen is 16, so molecular weight of O2 is 32 g. I didn't even look at a periodic table; I was thinking of sulfur. Sorry for any confusion.
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.
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
733966
So...find the balanced equation.\n\n2NO + O2 = 2NO2\n\nFind the masses of NO and O2.\n\nNO= 14g + 16g = 30g\nO2 = 2(16g) = 32g\n\nFind out the # of moles of NO you have...\n\n1mol/30g NO x 22.58g NO = .7527mol NO\n\nFind out how many moles of O2 will react with NO\n\n1 mol O2/ 2 mol NO x .7527 NO = .3763 mol O2\n\nFind out grams of O2 that will actually react...\n\n.3763mol O2 x 32g/1 mol O2 = 12.0427g O2\n\nAdd the amount of NO with the amount of O2 that will react...\n\n22.58g NO + 12.0427g O2 = 34.623g NO2\n\nI am NOT sure about this answer, I haven't done this in a long time....... Hope it helps.
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.
289562
First of all, it's spelled stoichiometry.\nSTP=22.4L per mole\nmethane = CH4\nmolar mass of methane = 4*1.0079(mass of 4H) + 12.011(mass of C)=16.0426gCH4\n\nSo it would be like this\n3.35L / 22.4L * 16.042g = 2.39923g methane\n\nWith significant figures, it would be 2.40g methane
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
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.
809019
The first material is not what you give but N2O4\n\nAssume you start with 1 mole of N2O4. After dissociation you have 1 - .1605 = .8395 mole\nWhen 1 mole of N2O4 dissociates it gives 2 moles of product. Hence the .1605 moles gives .321 moles NO2.\nTotal moles .8395 + .321 = 1.1705\nPartial pressure for each gas is moles of gas/total moles multiplied by the total pressure.\neg for N2O4 the partial pressure is .8395/1.1705 x 1 X 10^5 pa. \n\nTake it from here.
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
641455
Least C6H6 with like .5 moles\nThen 1 mole of CH4 and the 44 g CO2 both have same\nAnd then the 2 mole Br2
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2