the equation for the complete oxidation( complete combustion) of  c3 h8. what is the coeffcient in front of th
826680
C3H8 + 5O2 ---> 3CO2 +4 H2O\n\nThe coefficient for the O2 is    5
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
739541
CH4 + 2O2-> CO2 + 2H2O
780875
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
724209
Delta Hf = Delta Hf of the products - Delta Hf of the reactants\n\nThe Delta H's of pure species in their natural form are zero. That means that O2 and H2 will have delta Hf's of zero.\n\nThe other values you have to look up.\n\nOverall it will be...\n\nDelta Hf = 2*Delta H of CO - 2*Delta H of CH4\n\nThe twos are the stoichometric coefficients.
582466
1 goes in front of CO2.\n\nFe2O3 + CO ---> 2FeO + CO2\n\nAluminium chloride, AlCl3 because\nAl3+ --- group 3\nCl- --- group 7
133772
Any time you have carbon, hydrogen, and oxygen going through a chemical reaction they produce carbon dioxide (CO2) and water(H2O) \nSome are: C3H8(g) + 5O2(g) -> 3CO2(g) + H2O(g) AND\nC2H5OH + 3O2 -> 2CO2 + 3H2O\nTry to think of the last one on your own. Hope that helped.
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
394069
3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nFirst you must calculate the (average)oxidation number of central atom of main compounds in reactants.\nEx;Cr oxidation no=+6 in K2Cr2O7\n   C  oxidation no=-2 in C2H5OH\nThen in the products.\nEx;Cr oxidation no=+3 in Cr2(SO4)3\n   C  oxidation no=0 in CH3COOH\n\nThen you must evalute how many electrons must be transferred.\nEx;Oxidation no of Cr is changing from +6 to +3 \n   So it is reduced.\n   SO each atom of Cr must receive 3 electrons.\n   So totally 6 electrons must be received by K2Cr2O7\n   \n   Oxidation no of C is changing from -2 to 0\n   SO each atom of C must donate 2 electrons.\n   So totally 4 electrons must be donated by C2H5OH\n\nThen you must write now a explantion reaction.\nEx;you must write oppsite (simple)propotion of electrons in front of each compound\n  3C2H5OH + 2K2Cr2O7\n \nThen you can calculate how many Cr,C(,K) atoms should be there\n So, 3C2H5OH + 2K2Cr2O7=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nThen you can calculate the medium(SO42-) should be there\n i think 8 SO4 should be there\n\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nFinally you must equilize H,O atoms(balanced by water)\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nSecond one ................\nC2H5OH+4I2+6NaOH=CHI3+HCOONa\n                  +5NaI+5H2O
630069
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.
797059
118 gram of C3H8 is 118/(3*12+8) = 2.682 mol.\nThe reaction formula is C3H8 + 5 O2 -> 3 CO2 + 4 H2O, therefore 5*2.682 = 13.4 mol O2 is needed.\nAt STP this is 13.4 * 22.4 = 300 liter.\nSTP corresponds to 273 K and 101.3 kPa (760 torr).\nAs PV/T is constant, we can write\nV = (P_STP/P) (T/T_STP) V_STP = (100/101.3) (302/273) 300 =328 liter = 0.328 m^3.
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
802803
The answer is:\n2Al(HCO3)3  -------> Al2(CO3)3  +  3CO2  +  3H20\n\nThe equation is now fully balanced as you have\n2 moles of Al on each side\n6 moles of H on each side\n6 moles of C on each side\n18 moles of O on each side\n\nHope that helps!
936398
To understand this question, you need to know what kind of product will be produced during a combustion. Generally, for a complete combustion of a hydrocarbon (contains only H and C), the product will be H2O and CO2. Depending on the C and H content in the hydrocarbon, you will produce different amount of H2O and CO2. \n\nOn the other hand, if it is an incomplete combustion (with limited supply of O2), apart from CO2 and H2O, you may also produce C and CO. The relative amount between them varies depending on the amount of O2 that is made available. With very limited amount of O2, more C (soot) will be produced. This leads to the "sooty" flame you are seeing.\n\nIn order to explain why generally alkenes require less O2 to burn compared to alkanes, let's compare ethane (2 C alkane) and ethene (2 C alkene). \n\nEthane has a formula C2H6 while ethene has a formula C2H4. Their corresponding balanced equation for combustion are:\n\nC2H6 + 7/2 O2 → 2 CO2 + 3 H2O\nC2H4 + 3 O2 → 2CO2 + 2 H2O\n\nAs you see, since both compounds have same number of C, they both produce 2 mole of CO2 for every mole of compound. However, since they have different number of H, the number of moles of H2O produced will be different. Since H2O has a fixed formula, in order to produce it, you need to supply O2. That's why the two compounds require different amount of O2.\n\nThe reason that you see a sooty flame is due to the fact that you don't have enough oxygen supply during the burning process. You can easily confirm this by using just a Bunsen burner (but be very carefully!!!). First, turn on your Bunsen burner with the air hole closed. The flame you see at the moment should be a luminous flame. It is golden yellow color. Now, if you carefully put a glass plate (hold with clamp or other device) on top of the flame and slowly move towards the tip of the burner (but don't cover it completely), you should see the plate will get "darker" by the flame. This is the soot produced during the combustion. Now, take away the plate, open the air hole in the burner to the max. You should see a the flame turns to blue. If you put the glass plate (with the soot marks) back to the flame, you can see the flame can slowly "clean up" the plate (if enough O2 is supplied with the flame) and the soot will disappear. The soot is turn back to CO2 due to the excess of O2.
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
809514
Here you are:\n\nFirst write the combustion equation and balance it:\n\nH3NNH2(l) + O2(g) ---> N2(g) + 2H2O(g).\nThe heat will be the diffrence of enthalpies of formation between the products and the reactants. Do not forget to multiply by the stoichiometric coefficient.\n\nRemember that the enthalpies of formation of O2 and N2 are 0.\n\nYou will also need to find the enthalpy of formation of H2O(g) in the table of your book.  It is -241.8 kJ/mol\n\nHf(H2NNH2) = [2*(-241.8) + 1*0] - [1*50.6 + 1*0]\n\nHf(H2NNH2) = -483.6 - 50.6 = - 534.2 kJ/mol
684663
Maybe I'm wrong but I don't see a solution.  Not enough oxygen to complete that reaction.  Dod you have a typo somewhere?\nIf water is on the left then:\n  CH3COOH + 2 H2O + 8 NaCO3  =  8 NaHCO3 + 2 CO2