write the equation for the complete oxidation of c3h8. what is the?
954811
C3H8 + 5O2 --> 3CO2 + 4H2O
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
132050
oh shit.ok\nn2h2+(ch3)2n2h2+3n2o4 = 5n2 + 6co2 + 5h2
462663
C4H9OH + 6O2 --> 4CO2 + 5H2O
145111
The reaction formula is\n\nCH2COOH                           \nCH(OH)COOH (aq) + NaHCO3 (aq)----> \nCH2COOH                            \n\nCitric Acid\n\n       CH2COONa\n CH(OH)COONa    (aq)+ H20 (l) + CO2(g)\n       CH2COONa
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
228649
CO2 + H2O + light = C6H12O6 + O2
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.
565269
Complete oxidation of ascorbic acid would produce carbon dioxide and water.  The reaction would be:\n\nC6H8O6 + 5 O2 ===>  6 CO2 +  4 H2O
823470
First write the two half reactions:\n\n3 e(-) + Bi(OH)3 ---> Bi\nSnO2(=) ---> snO3(=) + 2 e(-)\n\nMultiply the first equation by 2 and the second by 3 and sum to get:\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi\n\n12 O in the left, 9 on the right, so add 3H2O\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi + 3H2O\n\nCheck that H is OK.  Yes. Done\n\nSo the sum is 13. (d) is the answer.
818293
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
754704
You should write NH4(+) otherwise you will not get the oxidation number right.  H is 1 so H4 is 4.\n\nIn order to be left with 1(+) you need 4 - 3 = 1.  So N will be -3.\n\nCheck -3 + 4*1 = +1 OK.
929785
You must figure out the oxidation number of each element in an equation.  The substance oxidized goes up in oxidation number (becomes more positive) and the substance reduced goes down in oxidation number (becomes more negative).  The substance that is oxidized is the reducing agent (causes reducing in the other element) and the substance that is reduced is the oxidizing agent (causes oxidation in the other element).\nSO . . .\nin (a) carbon on the left side is -4 and on the right side is +4 so it is oxidized and CH4 is the reducing agent\n     oxygen on the left side is zero (all elements are zero) and on the right side is -2 so it is reduced and O2 is the oxidizing agent\nin (b) zinc on the left is zero and is +2 on the right, so it's oxidized and Zn is the reducing agent\n     hydrogen on the left is +1 and is 0 on the right, so it's reduced and HCl is the oxidizing agent\nI'll leave the rest for you!
563324
There are a couple of methods but the one I like to use is the method of half reactions.  First we split the reaction into an oxidation half reaction and a reduction half reaction as follows:\n\nCr2O7(-2) -> Cr(+3)   and    CH3OH -> HCO2H\n\nWe know that the left half reaction is the reduction half because the oxidation number of the Cr changes from (+6) to (+3).  Likewise the right reaction is the oxidation because the oxidation number of the carbon changes from (-2) to (+2).\n\nNow that we have our half reactions the next step is to balance everything EXCEPT oxygen and hydrogen.  In this case the only other elements involved besides O and H are Cr and C.  Doing this give us: (the right one is already balanced for carbon; one on each side).\n\nCr2O7(-2) -> 2Cr(+3)   and    CH3OH -> HCO2H\n\nStep three: We balance oxygen be adding water molecules to the side with less oxygen atoms.  Doing this give us:\n\nCr2O7(-2) -> 2Cr(+3) + 7H2O   and   H2O + CH3OH -> HCO2H\n\nStep four: Since we are in acid we add H+ ions to balance the hydrogen atoms on both sides.\n\n14H(+1) + Cr2O7(-2) -> 2Cr(+3) + 7H2O      and         \n\nH2O + CH3OH -> HCO2H + 4H(+1)\n\nStep five: We balance the charges on both sides by adding electrons [e(-1)] to the side with the most positive charge.  Looking at the first half reaction we see that there is a total charge of +14-2 = +12 on the left side, and a total charge of 2*(+3) = +6 on the right side.  Therefore adding 6e(-1) to the left side should balance out the charges giving us: (do the same for the second half reaction)\n\n6e(-1) + 14H(+1) + Cr2O7(-2) -> 2Cr(+3) + 7H2O    and\n\nH2O + CH3OH -> HCO2H + 4H(+1) + 4e(-1)\n\nYou know you have made a mistake if the electrons don't end up on different sides of the two half reactions.\n\nStep six: We must balance the two half reactions to each other i.e. the number of electrons produced in one reaction must equal the number of electrons consumed in the other reaction.  If we multiply everything in the first reaction by 2, and everything in the second reaction by 3 this will give us 12 electrons in both half reactions.  Doing this gives us:\n\n12e(-1) + 28H(+1) + 2Cr2O7(-2) -> 4Cr(+3) + 14H2O    and\n\n3H2O + 3CH3OH -> 3HCO2H + 12H(+1) + 12e(-1)\n\nStep seven: Add the two half reactions together and simplify the result as much as possible by cancelling things that appear on both sides, and by finding the smallest set of whole number coefficients for the reactants and products.  Adding the half reactions together gives us:\n\n12e(-1) + 28H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H + 12H(+1) + 12e(-1)   \n\nCanceling the electrons on both sides gives:\n\n28H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H + 12H(+1)\n\nCanceling the extra H+ ions on both sides gives:\n\n16H(+1) + 2Cr2O7(-2) + 3H2O + 3CH3OH -> \n4Cr(+3) + 14H2O + 3HCO2H\n\nCanceling the extra H2O molecules on both sides gives:\n\n16H(+1) + 2Cr2O7(-2) + 3CH3OH -> 4Cr(+3) + 11H2O + 3HCO2H\n\nIt looks like this is the most simplified version.\nLast step: Check to see that the final reaction is balanced for both atoms and charges.  It looks like this is the final answer.
930613
CHEMICAL EQUATION\n\nC3H8 + 5O2 --> 3CO2 + 4H2O\n\n117g C3H8 * (1mol C3H8 / 44.09652g C3H8) = 2.65mol C3H8\n2.65mol C3H8 * (5 mol O2 / 1 mol C3H8) = 13.3 mol O2\n\n14.0°C + 273.15 = 287.15K\n\n 100.0 kPa (1 atm / 101.325 kPa) = .9896 atm\n\nPV = nRT\nV=nRT/P\n\nV= (13.3 mol*.0821L atm mol-1 K-1 * 287.15K) / (.9896 atm) = 317 L O2
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
961389
first, write down the reaction:\n 2C3H8 + 7O2 -> 6CO2 + 8H2O\n\nthen, convert gram of C3H8 to mol -> 160/44 = 3.64 mol.\nthird, find the number of mol needed to burn C3H8 by looking at the coefficient of the O2 and C3H8 in the reaction.\nn of O2 = 7/2 * 3.64 = 12.74 mol.\n\nafter getting mol of O2, we can compute the volume of O2 using:\nPV = nRT. remember that the temperature is in Kelvin not in Celcius. the pressure must also be in atm not in Pa.