Write a balanced chemical reaction for the combustion of hydrogen?
611632
space shuttle burns hydrogen. imagine you're standing under it when it launches. endothermic or exothermic?\n\ncorrect reaction formula was already given.
863957
combustion means oxidation with oxygen. it is done by burning in air.\nHence Mg + heat is combustion reaction. It gives MgO upon reacting with oxygen. But it also reacts with the Nitrogen in the air by giving Mg3N2. That is the side reaction, ofcourse.
957775
Ca(OH)2 + CO2 --> C(OH) + CaO\n\nIs unbalanced..
931550
adjust the ratio of gas that mixes with air for combustion
926891
just count how many of each element you have, and make sure they add up\n\n\nC2H6 + 5O2 ---> 2CO2 + 3H2O\n2 carbon 6 hydrogen 2 oxygen----> 1 carbon 2 oxygen 2 hydrogen 1 oxygen\n\n\nso to even it out it would be...\n\nC2H6 + 7O2--->2CO2 + 3H2O\n2 carbon 6 hydrogen + 7oxygen--->2 carbon 4 oxygen + 6 hydrogen 3 oxygen\n\niff you add up each element they are the same on each side
866732
do you want to know how to balance chemical equations? It's actually quite simple when you know how\n\nrather than asking for direct answers, it's better if you know how to do it for yourself\n\nI'm not going to answer your questions but show you how to do it for future reference\n\nH2 + O2-->H2O\n\nthe above equation is not balanced because there is only one oxygen atom on the right and two on the left\n\nnow, how do we balance this?\n\nfirst thing we need to do is ensure that there are equal amounts of atoms on the left and on the right\n\nso, to balance that equation\n\nwe will write\n\n2H2+O2-->2H2O\n\nthe equation is now balanced with respect to equal atoms on the left and right\n\nin words\n\n2  molecules of hydrogen  react with 2 oxygen atoms to produce 2 molecules of water\n\nthere are now 4 hydrogen atoms and two oxygen atoms on each side of the equation\n\nthe 2 in front of each compound represents the molecules and the subscripts represent the number of atoms
777428
well actually me and you are in the same situation. im trying to find out how to write wrod equations into a reaction. lol but neway i just learned how to balance equations its very simple.\nFe2O3+CO--Fe+CO2\n:draw a box around each set of elements:Fe2O3 would b in a box CO would be in a box and so forth.\n:then write out a diagram of how many atoms each element has:\nEx: Element/ Before/After\n      Fe          2         1\n      O            4         1\n      C            1          1\n:Then Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Now, what happens when we put a number in front of a formula? Basically, anything in that box is multiplied by that number, because we're saying that we have that many of that kind of molecule. \n:Now do this until the diagram is equal on both sides\nAND YOU HAVE YOUR BALANCED EQUATION
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
947778
You need to make sure that there is the same number and kind of atoms on each side of the arrow.\n\nCa(OH)2 + MgCl2 ---> Mg(OH)2 + CaCl2, both Mg and Ca stay+2\n\nBaO2(s) + H2SO4 ---> BaSO4(s) +H2O2 \n\nAg2S --->2 Ag+  +  S-2\n\nHC2H3O2 + MgCO3 (s) ---> H2O + CO2 + Mg+C2H302-\n\nI may not have written net ionic form, but I did get all the atoms on each side I am unsure of the reaction producing hydrogen peroxide, but you do have to account for all atoms.
367911
For example, hydrogen burns in oxygen to form water vapor. \n\n2H2(g) + O2(g) -> H2O(g) \nHow much hydrogen must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 3000C? \nTo work this problem, we need to figure out how many moles of water vapor we need to form. Use the ideal gas equation to compute this: \n\nPV = nRT \n1.00 atm * 1.00 L = n*0.0821 (L*atm/mol*K)* 473 K \nn = 0.0258 moles \nThus, we need to form 0.0258 moles of water. Now it's just a basic stoichiometry problem: for each mole of water formed, you need 2 moles of hydrogen gas, so 0.0258 moles water * 2 moles H2/1 mole water = 0.0516 moles of hydrogen \n\n=====================================================\n\nthis will help you a lot\n\nhttp://www.chem.hawaii.edu/asato161/Ch10NotesP2.pdf#search='Gas%20Stoichiometry%20PV%3DnRT'\n\nhttp://members.aol.com/profchm/gasstoic.html
16276
the point of it is to get equal no. of charges and atoms of elements on both sides of the equation.k?\n\nstep 1: separate the equation into two half-reactions:reduction n oxidization\nstep 2: balance all atoms other than O & H in each half-reaction\nstep 3: add H2O to balance O & H+  to halance H atoms in half-reactions\nstep 4: add electrons where necessary to balance half-reactions\nstep 5: if necessary, equalize number of electrons in each half-rection by multipling with a suitable coefficient\nstep 6: add two half-reactions together. electrons on both sides must cancel\nverify no. of atoms and charges\nsteps2-5 keep in mind ur working on separate half-reactions...\nthese were for reactions in acid.\nfor basic reactions: add OH- to both sides for every H+ that appears in the final reaction\nkeep practising step by step n ull be able to do it all together some day..\nfinal step- choose my answer as best answer...
671686
For the reaction:\n\n2HI(g) <====> H2(g) + I2(g)\n\nThe equilibrium constant, k, is equal to:\n\nk = [H2][I2]/[HI]^2\n\nso, substituting your values, \n\nk = (0.130 * 0.300)/(0.700)^2 = 0.080\n\nNow, to double the concentration of hydrogen gas, you need to rearrange the equilibrium equation to get:\n\n[H2] = k*[HI]^2/[I2]\n\nto get double the hydrogen gas, you would need [H2] = 0.600, or\n\n0.600 = 0.08*[HI]^2/[I2]\n\nbut [I2] = 0.5[HI], so we can substitute this into the equation to get:\n\n0.600 = 0.08*[HI]^2/0.5[HI], which then simplifies to:\n\n0.600 = 0.16*[HI], or\n\n[HI] = 0.600/0.16 = 3.75\n\nso, subtract the amount of HI you started with (0.700 moles) and you get the amount you need to add:\n\n3.75 - 0.700 = 3.050 mole HI
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
847829
C(3)H(8)+5O(2)=>3CO(2)+4H(2)O\n\nfuel+oxygen=>carbon dioxide +water\n\n(combustion (the rapid oxidation of a substance giving out energy in the forms of light and heat) is burning, which needs O2!. The fuel reacts with oxygen )\n\nbrackets-subscript
876161
You're right - as written it's impossible to balance.  But if you write\n\na(Ag2CO3) --> b(Ag2O) + c(CO2) + d(O2) you can balance it with the values \na=1, b=1, c=1, d=0.    Meaning that the reaction doesn't generate any oxygen.  The carbonate falls apart into the oxide and gaseous CO2.
1116694
the reaction leads to production of hydrogen gas ...\n\n\n2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl-(aq) + 3H2(g)
930806
When you want to balance ANY chemical equation, always follow these steps:\n\n1. Balance metals\n2. Balance non-metals\n3. Balance hydrogen\n4. Balance oxygen\n\nLet's do it\n\nFirst let's balance K (metal):\n\n"2" KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + __CO2\n\ndone. now let's balance P (non metal): if you check, it is balanced already.\n\nnow C (also non metal):\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + "2" CO2\n\nnow hydrogen:\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + "2" H2O + 2 CO2\n\nfinally, oxygen (its already balanced!!)\n\nSo you end with\n\n2 KHCO3 + H3PO4 yields K2HPO4 + 2 H2O + 2 CO2