For the reaction: CH4   +   O2   -->   CO2   +   H2O?
820077
First, balance the equation\n\nthe balanced equation should be\nCH4 + 2O2 --> CO2 + 2H2O\n\nnow convert 80.20g to moles. (i'm assuming you know how to do that)\n\nnow for every mole of CH4, you need two moles of O2.  So multiply the amount of moles by two.\n\nnow convert however many moles you have to grams O2\n\nand now you're done!!\n\nhope this helps!
391116
2Ca(s) + O2(g) -> 2CaO(s)
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
851681
Write the balanced equation - you need the mole ratios.\n\nChange 1g hexane to moles of hexane.\n\nSee how many moles of oxygen are USED and how many moles of  CO2 and H2O are formed.\n\nYour total number of moles in the reaction mixture would then be:\n\nmoles of Oxygen NOT used + moles of CO2 (we will assume the water formed consenses to a liquid at 20 degrees).\n\nUse the Ideal Gas Law to figure out the volume.\n\nPV = nRT\n\nwhere P = given pressure - water vapor pressure\nwhere V = your unknown\nwhere n = your total moles of gas\nwhere R = the appropriate Universal Gas Law constant\nwhere T = the temperature in KELVIN
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
780875
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
744647
C(3)H(8)+5O(2)=>3CO(2)+4H(2)O\nfuel+oxygen=>carbon dioxide+water\ncombustion \nbrackets-subscript
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
963470
CO + 2H2  --->  CH3OH\n\nmoles H2: 2.00g/ 1g/mol = 2.00 mol\nmoles CO: 18.50g / 28g/mol = 0.6607mol\n\nyour limiting reactant is CO since for every 0.6607 moles of CO, you'll only need 1.321 moles H2. there's an excess of 0.6790 moles H2.\n\ntheoretical yield, therefore, is (1.321/2)*100 =  66.05%\n\nbased from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,\n\nweight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams
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
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
133772
Any time you have carbon, hydrogen, and oxygen going through a chemical reaction they produce carbon dioxide (CO2) and water(H2O) \nSome are: C3H8(g) + 5O2(g) -> 3CO2(g) + H2O(g) AND\nC2H5OH + 3O2 -> 2CO2 + 3H2O\nTry to think of the last one on your own. Hope that helped.
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
835111
4 NH3  +  5O2  --->  4 NO  +  6H2O\n2 NO  +  O2 ---> 2 NO2  +  N2O4\n3 NO2  +  H2O ---> 2 HNO3  +  NO
550972
CO2 + H2O --> CH2O + O2\n read the attached webpage