The alcohol in 'gasohol' burns according to the following equation:?
916169
2.17 moles\n\nthe ratio is 3:1, not 2:1 as the above answerer states
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
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
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
1110418
C2H5OH\n\nThe end (-anol) tells you it's an alcohol, which means it has an OH group somewhere.  The begining (eth-) is the prefix used in organic chemistry to mean two, which tells you there are two carbons.  The two carbons are bonded to each other (that's just how alcohols work, because of the electronegativities of the atoms), and each one can bond to three other things.  If it were an alkane (the simple ones that are just a carbon chain surrounded by the maximum possible hydrogen atoms), it would be two carbons with six hydrogens around it (which is called ethane), but since it's an alcohol, there has to be an OH group somewhere, which is added basically by sticking an oxygen between one of the carbon-hydrogen bonds in the alkane.  That leaves you with two carbon atoms, six hydrogen atoms, and one oxygen atom, giving a molecular forumla of C2H6O.  But because the OH is basically one thing, it's generally written like that is a single atom, so the molecular forumla for ethanol is given as C2H5OH.
727216
use beers law to figure out the concentration and then take that to mole ratio
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
474531
CH3COOH + NAOH =.CH3COONA +H2O
462663
C4H9OH + 6O2 --> 4CO2 + 5H2O
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
947223
Alcohol, aldehyde, ketone, or carboxylic acid.
501588
Ether, or diethyl ether as it is more correctly called is made by combining ethanol molecules with each other on a catalyst so that ether and water is formed as the product. 2Ethanol + heat + catalyst = 1 diethyl ether + 1 H2O
351293
it is based on arrhenius equation\nK=A e ^ (-Ea/RT)\nA- frequency factor\nEa- activation energy (proportional to Delta H)\nR -Gas constant\nT -Temperature\nSince you dont know A- fequncy fator, yu need to do a lengthy calculation\n\nThe modified arrhenium equation will be ( by considering two situations T1, T2 there will be two rates k1, k2 but A-frequency factor is smae for one reaction so it's a constant..)\n\nEa= (R X T1 X T2 X Ln (k1/k2) ) / (T1-T2)\n\nEa= delta H = 17,390 J\nR = 8.314 J/K\nK1 = 1.6 X 10E-4\nK2= ?\nT1 = 298\nT2= 273+97= 370K\n  After substituing all these values, you will get K2= 6.27 E-4\nwhich is in agreement with theory (At high Temperatures, rate should be high..!)\n\nIs this info will help you?\nGood luck!
807508
Most of the info provided is superfluous.\nSimply put, you burned something and got 4.161g H2O (MW=18.016), all H coming from the sample.\nSo you formed 4.161/18.016 = 0.23096 mols H2O, and since each mol H2O contains 2 mols H, your sample contained 0.4619 mols of H.\n\nBTW, it is easy to deduce the empirical formula of the compound:\nNo hydrocarbon can have a MW of 60.095, and since only CO2 and H2O were formed, the compound  must contain O...this works out to be C3H8O, which means it is probably propanol (n or iso) or methyl ethyl ether.
74179
There are thousands of alcohols, its a class of organic compounds(made of carbon chains). All alcohols have a hydroxyl group attached somewhere on the carbon chain.\nThe general formula for alcohols is R-OH, where R is the carbon chain.
927756
2H2O2 → 2H20 + 02 is the balanced equation for the above decomposition
363124
detlaH=H(products) - H(Reactants)\n\n200KJ/mol