Balance the equation and calculate the enthalpy change for the following reaction:?
815531
It is simply sum of products minus sum of reactants\n\n[H2O + N2] - [O2 + NH3]\n\nSince this is a 1 to 1 ratio reaction, the coefficients are one. (if there was a coefficient for NH3, you must multiply the enthalpy value of NH3 by that # because that is how many moles you have, and the value given is for only one mole.)
890894
C12H22O11+12O2-------12CO2+11H2O+E
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
199900
The easiest way is to simply find literature listing delta H for the reaction.  Otherwise you have to do one of 2 things.  \n\n1. Find the enthalpy of formation for the product of the reaction which should be what you're looking for.\n\n2. Use the equation G=H-TS, plugging in the Gibbs free energy, temperature, and entropy values for the reaction,then solving for H, the enthalpy
867158
2Mg+O2-->2MgO\n2MgO+2H2o---->2Mg(OH)2
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
550972
CO2 + H2O --> CH2O + O2\n read the attached webpage
868500
Oxidation :\n\nCu > Cu+2 + 2e-\n\nReduction :\n\n2H+ + e- + NO3- > NO2 + H2O\n\nMultiply the 2nd equation by 2, you get the complete balanced redox equation :\n\nCu + 4H+ + 2NO3- > 2NO2 + 2H2O + Cu+2
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
970544
Here is what you do: \nDo the same operations as I did on the equations with the H values and you will be OK (Do not forget to use the same multipliers on the H as I did on the reactions).\n\n10PCl3 + 5O2 ---> 10Cl3PO\n10PCl2 + 10Cl2 --- 10PCl5\nSubstract the equations to get (keep coeff positives by moving to the other side):\n\n5O2 + 10PCl5 ---> 10Cl3PO + 10Cl2\nP4 + 5O2 ---> P4O10\nSubstract and get\n10PCl5 + P4O10 ---> P4 + 10Cl3PO + 10Cl2\nP4 + 6Cl2 ---> 4PCl3\n\nSum the two to get (P4 goes):\n\n10PCl5 + P4O10 ---> 10Cl3PO + 4Cl2 + 4PCl3\n\n4 PCl3 + 4Cl2 ---> 4PCl5\n\nSum the two equations\n\n10PCl5 + 6P4O10 ---> 10 Cl3PO.
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
927756
2H2O2 → 2H20 + 02 is the balanced equation for the above decomposition
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
976256
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)