how do you do theoretucal and percent yield and find out limiting reactant in the simplest way?
543827
If you know the grams, which you do on one...its easy.  To find the grams from ml, you need to know the density.  Density is in grams/ml, so if the density was 2 g/ml and you had 4.5 ml, then you would have 9 grams.  So you take the grams and you need to find moles to calculate percent yield.  To do this, find the total molecular weight (for instance, water is 18 b/c of two H's which are 1 and one O which is 16).  When you have the grams and molecular weight, divide the grams by the molecular weight.  This will give you the moles, since the molecular weight is grams per mol.  Do this for every reagent or thing you use.  Whicever one has the lowest amount of moles is your limiting reagent because once that is gone, the reaction stops!  Now the theoretical yield is found by doing stoiciometry.  So whatever your limiting reagent is...that number is your theoretical yield for this problem.  Then weight the product when you are done, and find the amount of moles you have by oncve again dividing how many grams you have by the molecular weight.  Then divide that by your Theoretical yield and that is your percent yield!  Looks harder than it is.
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
805472
The first step in working with limiting reagents problem is to balance the chemical equation.\n\n2 Al (s) + 6 HCl (aq) ====> 2 AlCl3 (aq) + 3 H2 (g)\n\nNow calculate the number of moles of each reactant\nMoles of Al = 14.6g / M.W. of Al\nMoles of Al = 14.6g / 27g/mol = 0.5407 moles\nMoles of HCl = 14.6g / M.W. of HCl\nMoles of HCl = 14.6g / 36.45g/mol = 0.4005 moles\n\n1.        Look at the coefficients of the reactants in the balanced equation. For every 2 moles of Al  6 moles of HCl is required or for every 1 mole of Al 3 moles of HCl is required\nIf you used up 0.5407 moles of Al then 3 * 0.5407 (1.6621) moles of HCl will be needed. Since only 0.4005 moles of HCl were used instead of 1.6621 moles, therefore it is the limiting reagent.\n\n2.\nMass of AlCl3 = Moles of HCl * coefficient of AlCl3 / coefficent of HCl * Molecular weight of AlCl3\n\n3.\nMass of H2 = Moles of HCl * coefficient of H2 / coefficent of HCl * Molecular weight of H2
942137
I have two problems with this: First, this is not an avenue for high school kids to get their homework done for them and second, the thought that someone would take yahoo answers 'points' as payment for doing a highschool kids homwork makes me feel ill. \n\n(i)All that aside, the reducing agent is the compound/element that looses electrons. \n\n(ii) First balance the equation, then find then number of moles of potassium permagenate used in the titration, then convert to the equivalent number of mols of the Be complex. \n\n(iii) multiply answer for (ii) by 50\n\n(iv) use the difference between your asnwer for (iii) and the given sample mass to find mass percent.
816363
4 Fe(s) + 3 O2(g) ----> 2 Fe2O3(s)\n\nFirst, you have to check if the chemical equation is balanced. In this case, it is. So, what it tells you is that for every 4 mole of iron that is reacted, 1652 kJ will be released to the surrounding. Now, since O2 is in excess, iron will then be your limiting reagent. If you have 3.34 mole of iron, you can use ratio to determine how much energy will be released.\n\nE = (-1652/4) *3.34 = -1379.42 kJ\n\nNote that the "-ve" sign indicates energy is released instead of being absorbed.
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
758495
The balanced equation for this reaction is :\n4Fe + 3O2 --->  2Fe2O3\nOnce you have the balanced equation, you can solve for the limiting reactant.\n0.45 mol Fe * (2 mol Fe2O3/4 mol Fe) ^^this is the mole ratio, derived from the balanced equation^^  =  .225 mol Fe2O3\nNow you have to do the same exact thing with 0.30 mole O2, your other reactant:\n0.30 mol O2 * (2 mol Fe2O3/3 mol O2) ^^once again derived from the balanced equation^^  = 0.20 mol Fe2O3\nThe limiting reactant in this reaction is O2 because it produces less moles of Fe2O3 than the Fe.\nYou use the Fe2O3 moles produced by the limiting reactant because that is all that can be produced with the "limiting" amount of reactants.  In other words, the answer is 0.20 moles of Fe2O3 are produced from the reaction.\n\nHope I helped..
949367
First you have to get the molecular weight of those.\n\nits\n180 for C6H12O6\n192 for 6O2\n264 for 6CO2\n108 for H2O\n\na. Since you have the same amount of C6H12O6 and O2, your limiting reagent in this case is the O2.\n\nSo you use it to determine the CO2 produced.\n\nb. if 192 grs of O2 produce 264 grs of CO2\nthen 100 grs will do: (100 * 264)/180 = 137.5 grs of CO2.\n\nc. Now, to calculate the percent yield, in ideal conditions you would have gotten 137.5 grs (which is your 100% theorically) but practically you got 97.1 grs. so\n\n137.5 ___ 100%\n97.1______ ?\n\n(97.1*100)/137.5 = 70.62% is the perecnt yield of CO2.\n\nd. You do the same thing backwards to determine how many excess of the non-limiting reagent (C6H12O6) you have.\n\nto produce 264 grs of CO2 you need 180 grs of C6H12O6\nfor 137.5...\n\n(137.5*180)/264 = 93.75 grs of C6H12O6\n\nso the excess is 100 - 93.75 = 6.25 grs of C6H12O6
385547
Percent Yield= (Actual yield divide by theoretical yield)*100\n\n\nsolve for Actual yield:\n\n\npercent yield= 92%/100=.92\n\n\n  molar mass of WO3 = 231.85g/mol\n\ntheoretical yield =   3.72g divide by 231.85g/mol = .016mol WO3\n\n\nActual yield = X\n\n\nx =  .016mol WO3*.92\n\n= .0147mol  WO3
955797
Use the Arrhenius equation to predict the rate:\n\nhttp://en.wikipedia.org/wiki/Arrhenius_equation\n\nK = Ae -E/RT\n\nIt all boils down to your reactants, the activation of energy, and the product that is being synthesized.
917127
Here's the how you do it.\n\nBalance the equation.  Element by element.\n\nThat gives you the mole ratios of  the various compounds.\n\nConvert the moles to mass using the molecular weight of each of the compounds.\n\nThat gives you the mass ratios of the various items.\n\nDivide both sides of the equation by the largest mass on the left side of the equation,    Then  you will have the mass of products per Kg of the at least one of the reactants.  Remember  you are starting with 1.0 kg of one item and 0.98 of the other.\n\nso you must check to see which reactant will limit the production.\n\nShould be a simple matter of working out the ratios.
978323
*your equation is written wrong it should be:\n3Cu(NO3)2 + 2Na3(PO4) -> Cu3(PO4)2 + 6Na(NO3)\n\nyou have to find your limiter and your excess.\n 1.) change grams to moles using moler mass\n2.2 g Cu(NO3)2 times 1mol/188 g = .0117mol\n1.29 Na3(PO4) times 1mol/164 g = .00787 mol\n\n2.) divide the two by the # of moles in the balanced equation:\nCu(NO3)2: .0117 mol/3 = .00390 mol\n Na3(PO4): .00787 mol/2=.00394 mol\n\n3.) So now you know you don't really have a limiter, but we'll go with Cu(NO3)2 b/c there's a little less. \n\nthere are 3 moles of Cu(NO3)2 used for every 1 mol of Cu3(PO4) produced.\n\n.0117 mol Cu(NO3)2 times (1 mol Cu3(PO4) / 3 mol Cu(NO3)2) = .00390 mol Cu3(PO4) produced\n\nconvert that to grams\n\n.00390 mol Cu3(PO4) times 286 g/mol = 1.12 g \n\nthat's your actual\n\nso % yield = .89/1.12 x 100 = 79.5%\n\noh yeah you find moler mass by adding the molecular weights of the atoms in the compound (times the number of the atoms in that compound) for ex. Cu3(PO4) 63.6 x 3 + 31.0 + 16.0 x 4 = 286\n\nDon't forget significant figures!!\n\nhope that helped!
608949
cant answer secifically, but you should be able to back calculte taking elements present today and using the half lives as your main variable.
270274
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...
749681
find the stoichiometric yield.  then multiply by .8962 voila.
363124
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
250561
i can help with part of it. i forgot a lot of my chemistry stuff. so ok  first thing is to find the number of mols of each reactant put in. which is  mass over molar mass. and then find the molar mass of 0.566g of H2. so you know how many moles of H2 are left over, so you should lknow how many moles reacted and you should know the number of mols of the other 2 substances too and you know the volume and you can figure out the masses so then   do that. \nand Kc is just the  conc of products over reactants. if youre in AP   i assume you know how to do that\nthen c   and    um partial pressures. i think it has something to do with mole ratios and   the total pressure. im not completely sure, but im sure you can figure it out. good job