How many units of propane are needed to balance this reaction?
552184
C3H8 + 5O2 = 3CO2 + 4H2O\n\n1 mole of propane is required.\n\nCheck:\nC: 3 on both sides\nH: 8 on both sides\nO: 10 on both sides
816907
It varies with the grade produced.  This example is is 94.5 nitromethane with nitroethane and nitropropane.  I doubt there is any measurable amount of propane left in the commercial product.
957905
the balanced reaction\nC12H22O11 + 12 O2 ---> 12 CO2 + 11 H2O\nso the right answer are:\nc. \nand\ne.
194284
To solve this problem we need to think about limiting reactants.  This one is fairly simple because I already have moles and the ratio is 1:1:1.  If I only have 3 kmols of C, that is the most I can make, because going beyond that, I would be out of C.  Thus, I can only make 3kmols of CO2 with 2 kmols of O2 left unreacted.  These amounts are theoretically find for a combustion reaction.
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
485922
the minimum amount of energy required to start a reaction
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
691513
Check that both sides of equation match. Here is an example:\n\n3 CH3CH2OH + 2 Cr2O7 2- + 16 H+ --> 3 CH3COOH + 4 Cr3+ + 11 H2O\n\nThere should be an equal amount of H, C, O, Cr in both sides of the equation. \nIt is sometimes difficult to know where to start balancing an equation. In such case you should try to find an element/molecule that does not appear in every molecule on the other side of the equation. Eg. when balancing this equation above, I first balance the number of Cr because it is on its own on the right side. Sometimes you just need to try and play with numbers for a while to find the right answer. Hope this helped...
122241
Are you sure you did not make a typo?\nA Hydrocarbon with the formula C7H17 would be very hard to find.  Usually, the number of Hydrogens in a ["normal"] Hydrocarbon can be determined by doubling the number of carbons and adding two.\nIn the case we have something that looks like Heptane (7 carbons), so I would expect there to be 16 Hydrogens (7*2 + 2).  If you align all the carbons in a row and single bond them to eachother to allow for the maxmium number of C-H bonds, where would you place the 17th Hydrogen atom?\n\nIf you did mean C7H17, then the balanced reaction would be,\n4 C7H17 + 45 O2  =  28 CO2 + 34 H2O\n\nHowever, if you did indeed mean Heptane (C7H16), the balanced reaction would be,\nC7H16 + 11 O2  =  7 CO2 + 8 H2O
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
1010084
Propane  = C3H8. The molar mass is\n(3 x 12 + 8 x 1) gram = 44 gram.\nYou give 44.1 g. That a little bit better.\n\nThe volume of the cylinder is\nAh = pi r2 h = 3.14 x (0.050 m)^2 x 1.30 = 0.212 m3.\n\nBefore: pV = nRT\n1.4×10^6 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(begin) = 119 mol.\n\nAfter: pV = nRT\n2.4×10^5 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(end) = 20 mol.\n\nUsed is (119 - 20) mol propane, being 99 x 44.1 gram = 4.4 kg
136369
reverse reaction one(and change the sign of delta H), add the two reactions together, cancel out on both sides. then add the delta H for both reactions, should get equals 54.08kJ
908510
It should be Fe2+ + 2OH- --> Fe(OH)2 since you not only need to balance the elements but you also need to balance the charges.
1012524
2C8H18 +   25O2  --->   16CO2 + 18H2O\n\n2 molecules of octane (which is not the only component of gasoline, and octane rating has nothing to do with how much octane is actually present) react with 25 molecules of oxygen to produce 16 molecules of carbon dioxide and 18 molecules of water.
930204
First, convert 16 g CH4 into moles.  Given the molecular weight of methane (CH4) is 16, we have one mole of methane on our hands.\n\nNow, find the equation for the reaction.  This is:\n\nCH4 + O2 -->  CO2 + 2 H2O\n\nIf we assume complete combustion of the CH4 (given excess oxygen), then we have one mole of carbon dioxide and 2 moles of water on our hands.\n\nFinally, convert that back to weight using the molecular weight of carbon dioxide (44) and water (18) and you get\n\n44 + 2 (18) = 80 grams of total products.
714714
A balanced equation for this combustion is:\n\nC3H8 + 5 O2 ----> 3 CO2 + 4 H2O\n\nSo, every mole of C3H8 requires 5 moles of O2 to burn.\n\n# of moles of C3H8 = 138 / (12*3+8) =3.136 mole\n\n# of moles of O2 required = 3.136*5 = 15.682 moles\n\nUsing ideal gas law, PV = nRT, we get\n\nV = nRT / P = 15.682*8.314*(25+273) / (100*1000)\nV = 0.389 m^3
250273
Enough to cancel out the H's. For the reaction with 2H's, add 2OH's to both sides. For the reaction with 3H's, ad 3OH's to both sides.\n\nH + OH --> H2O\n\nEDIT: Oh, well in that case, the H's cancel out, so you have just one H on the side that previously had 3. Then, just add one OH to both sides.