Write a balanced equation for the combustion of propane, C3H8.?
818293
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
729249
first balance equation\nCH4 + 2O2 --> CO2 + 2H2O\n\nthen convert (look at balanced equation and use coefficients)\n3.28 moles O2 x 1 Mole of CO2 / 2 Moles O2 = 1.64 moles of CO2
694411
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
165982
Propane - C3H8 = 44 grams/mole\nButane - C4H10 = 58 grams/mole\n1 cc = .001 liter\ngas constant R = .08206\n\nA)\n\nPV=nRT\n(1 atm)(50x.001L) = n(.08206)(273K)\nn=.002 mol\n\n.002 mol propane x (44 g/mol) = .088 g\n.002 mol butane x (58 g/mol) = .116 g\n\n50% propane by mole\n43% propane by mass\n\nB)\n\n(150 kPA)/(101.325 kPA) = 1.48 atm\nPt = P1 + P2\n1.48 atm total = .74 atm propane + .74 atm butane\n\nC)\n\nThe propane should effuse faster because its molecular mass is lower.
418645
heat of combustion = -(enthalpy change for reaction )\n          \n = -(heats of formation of products - reactants)\n \n1 mole of ethanol, when burnt completely, produces 2 moles of CO2 and 3H2O i.e delta H = 2(heat of formation of CO2) + 3(heat of formation of H2O ) \n\n= 2(-394 kJ/mol ) + 3(-286 kJ/mol)\n\n= -1646 kJ/mol\n\nheat of formation of ethanol = -277 kJ/mol\n\nso\n\nheat of combustion = - (-1646 - (-277) )\n                   \n                   = - (-1646 + 277 )\n\n                   = 1369 kJ mol
962190
C2H4O + 5/2O2 ----> 2CO2 + 2H2O, or get rid of the fractions to get:\n2C2H4O + 5O2 ----> 4CO2 + 4H2O
929071
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
195778
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
887756
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
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
905654
Propane is higher.  It's because C3H8 has a higher density than CH4.  There's a formula in "Perry's Chemical Engineer's Handbook" that lets you estimate the heat value of a hydrocarbon fuel, based on its density, which I don't have available right now...however, the following should be useful:\n\nhttp://cta.ornl.gov/bedb/appendix_a/Heat_Content_for_Various_Fuels.xls
826680
C3H8 + 5O2 ---> 3CO2 +4 H2O\n\nThe coefficient for the O2 is    5
278240
C6H12O2 + 6O2 --> 6CO2 + 6H2O\n\nis the combustion reaction for glucose.
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
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!
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.
954811
C3H8 + 5O2 --> 3CO2 + 4H2O