Please help me with balancing this equation (CH3)2 N2H2+N2O4-->N2+H2O+CO2?
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
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
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
59751
Write the balanced reaction.  From that write the equilibrium constant equation.  Plug in the starting numbers plus/minus the change and solve for x.\n\nClue - if the loss of N2 and O2 are x, the formation of NO will be 2x.
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.
986618
The equation is already balanced. Count the Ag's on each side = 1. Count the Na's on each side = 1. Count the Cl's on each side = 1. Count the NO3's on each side = 1. One can count the N's on each side = 1, or the O's on each side = 3.\n\nBut one could also consider the question another way:\nAg+ + NO3- + Na+ + Cl- ===> AgCl + Na+ + NO3-
774815
CH4 + 2 O2 ---> CO2 + 2 H2O
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
550972
CO2 + H2O --> CH2O + O2\n read the attached webpage
825261
First you must balance the equation:\n4NH3 + 5O2  -> 4NO + 6H2O\nthere are 4N, 12H and 10O on each side\nyou have 4.78 moles of O2 you need to divide it by the number of moles of O2 that react and multiply it by the number of moles of NO produced.\n4.78 / 5 x 4 = 3.824moles\nHope this helps
119628
So, usually the best way to do these is to just fiddle with them until you get them right, but for the sake of it, let's go with a systematic approach using algebra.  First we assign a variable in front of each of the terms on the left and right like:\n\nx Pb  +  y H2O  +  z O2  -->  a Pb(OH)2\n\nSo now we'll write down equations to make sure that we have the same numbers of both sides of each *element*.  I'll do this by writing down the variable in front of each compound in your equation (so x, y z and a) and then multiplying those by the number of times an element exists in that compound.  (So when I'm writing the equation for hydrogen, the H20 term will be 2*y, because y is the coefficient in front of H20 and H occurs twice in that compound).\n\nRemember, the number of any element on the reactants (left) side equals the number of that element on the products (right) side.\n\nFor Pb: x = a, because Pb shows up only once on each side with a subscript of 1, those two compounds will have to be present in the same molar amounts.\n\nFor Hydrogen: 2y = 2a\n\nFor Oxygen: y + 2*z = 2 * a\n\nSo now we have three equations:\n\nx = y\ny = a\ny + 2z = 2a\n\nBut four unknowns, which is fine because we'll just solve for them in terms of one of the variables and then simply choose the last variable to make all the numbers nice.  So solving the equations in terms of a:\n\nx = a\ny = a\nz = (2a - y)/2 = (2a - a)/2 = a/2\n\nSo now we have x, y and z in terms of a free parameter, a.  We'll just choose a = 2, for example, to make x, y and z all turn out as integers.  So now we've chosen:\n\nx = 2\ny = 2\nz = 1\na = 2\n\nSo the balanced equation is:\n\n2 Pb + 2 H2O + O2 --> 2 Pb(OH)2\n\nMake sense?
539827
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
835111
4 NH3  +  5O2  --->  4 NO  +  6H2O\n2 NO  +  O2 ---> 2 NO2  +  N2O4\n3 NO2  +  H2O ---> 2 HNO3  +  NO
775000
2 MnO4-(aq ) + 5 H2C2O4(aq ) + 6 H3O+(aq ) --> 2Mn2+(aq ) + 10 CO2(aq ) + 14 H2O
683249
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
750464
My chemistry is pretty rusty, but let's see.\n\nI'll assume that you have an infinite amt of O2, and that you have 68.12 grams of NH3.\n\nThe equation isn't balanced, so you need to do that first (you have 3 hydrogens on one side, but two on the other).  Balance to get:\n\n2(NH3) + (2.5)(O2) --> 2NO + 3(H20)\n\n4(NH3) + 5(O2) --> 4NO + 6(H20)\n\nNow, given that you have 68.12 grams of NH3, you need to convert this into moles of NH3, as Stoichiometry deals with molar ratios.  To get the conversion ratio, look at the periodic table.  Under each element, there should be an atomic mass.  For oxygen is 16.something, Hydrogen 1.008 ( i think ).  Find it for N and H.  (this is where I get hazy on the math).  I believe: take 3 * atomic mass of hydrogen + atomic mass of nitrogen, let's call it M  (it should be in grams/mole)  So to get moles, multiply 68.12 * (1/M).  \n\nNow, our equation above shows that for every 4 moles of NH3, we get 6 of H20.  # moles NH3 * (6/4) = # moles of H20.  Now, do the reverse operation of H20 to get how many grams is produced.  H20 has an atomic mass of something like 18.016.  So, # moles H20 * 18.016 = # grams H20\n\n\nI hope that's somewhat correct, and if not, leads you in the right direction at least