CH3OH + O2 -> H2O + CO2 + Heat\n\n(Methanol + oxygen -> Water + carbon dioxide + heat)\n\nMajor thank-you's to all those who will answer my question! The more, the better! :)
123069
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
15982
Simple one will be \nSunlight + Carbon dioxide (CO2) + Water (H2O)--> Glucose (C6H12O6) + Oxygen (O2).\nor \n6CO2 +12 H2O (light energy and chlorophyll) ---> C6H12O6 + 6O2\nRemember to use subscript whenever necessary.
18990
Photosynthesis\n\n6H2O + 6CO2 ----------> C6H12O6+ 6O2
167487
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
78615
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
302451
NaOH (Base) + HNO3 (Acid) -> NaNO3 (Salt) + H2O (Water)\n   H  H H\n   |  | |\nH-C-C-C-H\n   | | |\n   H H H\n\nReplace the last H by an -O-H\n\nDraw the carbon skeleton first.
403066
2Al + 3F2 ----> 2AlF3\nAl + 3OH- ----> Al(OH)3
397393
Reactants: Hydrogen and Oxygen gas are used up\nProducts: water is produced\n\nH2(g) + O2(g) -----> H2O (l)\n\nBalance O: H2(g) + O2(g) -----> 2H2O (l)\n\nRebalance H: 2H2(g) + O2(g) -----> 2H2O (l)
139991
decomposition reaction\n\n2H2O2 -------- O2   +  2H20
505986
2C6H14 + 19O2 ----> 12CO2 + 14H2O\n---or--\nC6H14 + (19/2)O2 -----> 6O2 +7H2O
533955
the balanced reaction\nC12H22O11 + 12 O2 ---> 12 CO2 + 11 H2O\nso the right answer are:\nc. \nand\ne.
402503
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
371445
CH4 + 2O2-> CO2 + 2H2O
375486
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
509333
(1) NaHCO3    --->  Na(+) + HCO3-\n(2) Ca(NO3)2  ---> Ca2(+) + 2NO3-\n(3) Al2(CO3)3  ---> 2Al(3+) + 3CO3(2-)\n\nThis is all taking place in water (ideally), so you should consider that too.
449333
This is very simple\n\nCa+H2O -> Ca(OH)2 + H2\n\nThe reactant has\n1 atom of Ca\n1 atom of H\n2 atoms of O\n\nThe product has\n1 atom of Ca, ok\n2 atoms of H\n2 atoms of O\n\nJust put a coefficient of 2 in the water (H2O)\n\nSo the chem. eqn becomes\n\nCa+2 H2O -> Ca(OH)2 + H2\n\nThat's it.\nThe equation is already balance
377713
1  This is obviously an addition reaction.  \nAl2 O3\nAl has a charge of 3+\nO has a charge of 2- , so O2 has a charge of 4-\nFind the lease common multiple, we want the charges to cancle when they combine and = 0\nSo lets just guess and check to find the charge\nAl+O = +1\nAl+O2 = -1......  Lets increment the Al up\nAl2 + O2 = -2  (  +3 * 2  + (-2) * 2)\nAl2 + O3 = 0.... so we find our solution is Al2O3\nThe balanced reaction is\n2Al + 3/2 O2 ----> Al2O3\n\n\n2\nCaO + H2O.....think about what type of reaction this is.  Is it a substitution?.   You'll just have the O's swap and nothing happens.   It must be an addition reaction\nCaO +H2O -->  Ca(OH)2   \nAgain balance the charges\nLooking at the periodic table\nCa is 2+\nWe know we are going ot get an OH.  WE could hypothetically get (OH)...(OH)3..etc.....  How do we know its (OH)2?\nOH most people know has a 1- charge.  But we can divide it up\nO is 2-...... H is 1+.   = 1-\nWe need 2 OH to make the 2+ charge of Ca cancle.  \n\n\n3 Same thing as was done in in question 2...it has to be an addition reaction\nH2O + CO2  --> H2CO3     \nCO2 generally dissolves in water.  (carbonated beverages)\nAnd only a small amount actually ends up as H2CO3.  The rest is just CO2 molecules floating in water\n\nI'll let you try to figure this one out..... You know you are going to one final molecule with all those atoms .  Figure out a molecule that makes the charges cancle.  THen just balance