could someone please tell me what the balanced equation of butanol in combustion is???\n\nThanks a lot this coursework is really pissing me off now :(
209188
C4H9OH + 6O2 --> 4CO2 + 5H2O
51199
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
463459
In a closed vessel the change in partial pressure for each gas during reaction scales to the stiochiometric coefficients.\n\nSo for this reaction, if 23 torr/min of NO is being used, then there must be 23 torr/min of NOCl appearing since both coefficients are 2 and they are on opposite sides of the equation.\n\nLikewise, if 23 torr/min of NO is being used, then 11.5 torr/min of Cl2 is also being used since they are on the same side of the equation.  It is only 11.5 torr/min because the coeffient for Cl2 is only 1, so it is half the amount of NO.\n\nThe total pressure change will be the sum of the amounts:\n\npNO + pCl2 + pNOCl = -23 -11.5 + 23 = -11.5\n\nKeeping two sig figs, this would mean that the pressure is decreasing by 12 torr/min.
420597
Ca + 2H2O yields Ca(OH)2 + H2\n\nBasically it's almost balanced already, just add a 2 in front of the H2O.
120806
Unless a equation is properly balanced the amount of product and reactant will not be the same, I suppose this is like a recipe where u get the proper number of ingredients(moles) and mix to together to give the you the proper number of products.
493280
I have a serious feeling that you haven't given us all the info.\nIf you know the density of HCl you can get the first one.\nBut we don't have enough info to even begin to tell you the relevance of the NaOH.   More info, or concentrate on your own homework...
511565
2H2O2 → 2H20 + 02 is the balanced equation for the above decomposition
332489
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
429913
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
505326
its actually pretty simple. U multiply by -1 when u flip over an equation...so the reactants become the products and the products the reactants. that way instead of having a negative number u will have a positive number for the kJ. as in this problem. U have to flip #1 so u can cancel a few things with #2. so it becomes..\n1)3 H2O(l) + 2 CO2(g) AH---->CH3CH2OH(l) + 3 O2(g) (multiply by -1 gives u a postive +1367kJ. part two stays the same.\n2) C2H4(g) + 3 O2 ---> 2 H2O(l) + 2 CO2 (g) AH= -1411 kJ\n\nNow the 3 O2 from #1 and #2 cancel since they are on opposite sides. Now the 2 H20 cancel from #2 and you are left with only one in #1 since they are also on opposite sides...which after canceling..leaves u with your original equation.\n\nC2H4(g) + H2O(l) ------> CH3CH2OH(l)\n\nnow since u flipped over #1.....and u have a positive +1367kJ and\n#2 is left the same...its still at negative -1411kJ. u add them together and u get an enthalpy change of -44kJ. sorry its pretty confusing to explain..if anything message me and i willl try to help.
466010
do you want to know how to balance chemical equations? It's actually quite simple when you know how\n\nrather than asking for direct answers, it's better if you know how to do it for yourself\n\nI'm not going to answer your questions but show you how to do it for future reference\n\nH2 + O2-->H2O\n\nthe above equation is not balanced because there is only one oxygen atom on the right and two on the left\n\nnow, how do we balance this?\n\nfirst thing we need to do is ensure that there are equal amounts of atoms on the left and on the right\n\nso, to balance that equation\n\nwe will write\n\n2H2+O2-->2H2O\n\nthe equation is now balanced with respect to equal atoms on the left and right\n\nin words\n\n2  molecules of hydrogen  react with 2 oxygen atoms to produce 2 molecules of water\n\nthere are now 4 hydrogen atoms and two oxygen atoms on each side of the equation\n\nthe 2 in front of each compound represents the molecules and the subscripts represent the number of atoms
151380
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!
538118
CO + 2H2  --->  CH3OH\n\nmoles H2: 2.00g/ 1g/mol = 2.00 mol\nmoles CO: 18.50g / 28g/mol = 0.6607mol\n\nyour limiting reactant is CO since for every 0.6607 moles of CO, you'll only need 1.321 moles H2. there's an excess of 0.6790 moles H2.\n\ntheoretical yield, therefore, is (1.321/2)*100 =  66.05%\n\nbased from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,\n\nweight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams
77932
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.
329410
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
381962
It is perfectly permissible to write the products as Br2 + Br2.  It is probably easier to work this way.  After you balance the equations, just add the Br2's together to give the final reaction
111731
ur equation is wrong....\nu will get H2O + C...\ncheck again\nif it's correct, u have to find out which one is the limiting reagent by calculating the number of moles of each reactant used...thus figuring out the number of moles of product and multiplying with the molecular mass of methanol...