Product yield production and solution stoich???? does anyone know anything? i have homework to do!!?
406601
When performing a chemical reaction you want to use as little of each material to generate the quantity of material you need, with as little waste as possible.\n\nFor example:\n\nHCl + NaOH -> NaCl + H2O\n\nYou are using 1 mole of HCl and 1 mole of NaOH to generate 1 mole of salt and 1 mole of water, therefore stoichiometryt is 1:1.  This is the stoichiometry.  \n\nThe reaction:\n\nH2SO4 + 2NaOH -> Na2SO4 + 2H2O\n\nThe stoiciometry is 1 mole of H2SO4 and 2 moles of NaOH, therefore 1:2\n\nStoichiometry is the relative amounts of each reagent you use in a reaction.\n\nIf you were to use:\n\n2HCl + NaOH -> NaCl + H2O + HCl\n\nYou are using 2 moles of HCl and 1 mole of NaOH to generate the same thing, but you have an excess of HCl left over.  This is yield.  The yield of this experiment in relation to HCl would be 50%, since you have only consumed half of it, the rest is waste.  With respect to NaOH the yield is 100% since you have used all of it.  The yield of the reaction though is 100% as it is always based on the limiting reagent.\n\nThe product yield is the amount of product you recover based on a theoretical maximum.  In the first example above, and assuming you did the reaction on a 1 mole scale, you should recover 1 mole of NaCl (57.5g) and 1 mole of H2O (18g).  If you only recovered 30g of NaCl, then the yield of the reaction is 30 / 57.5 = 52.2%.\n\nNo chemical reaction is perfect so getting 100% return is rarely possible, although we do get very close occasionally in industry.\n\nEcologically the less EXCESS of reagent you can put into a reaction to maximise the recovery limits the amount of material going to waste, which has an impact on the environment.  Economically the less EXCESS of reagent you need, means it costs less to prepare a compound.\n\nWhen pharmaceutical companies make drugs they spend millions on materials.  Cutting down by 10% how much of everything they need can save hundreds of thousands of pounds per year / per drug on excess materials expenditure, increased environmental exposure etc.  Chemical companies spend 100's of thousands of pounds a year disposing of excess chemicals safely, recycling chemicals etc and are liable to very heavy fines if they don't.  Many savings can be made just by improving stoichiometries and product yields.\n\nLast year I spent over $1 million on chemicals, by improving stoichiometries and yields by 10% this alone saves an enormous amount of money.\n\nI hope this helps?
951720
BaCl2 + ZnSO4 => BaSO4 + ZnCl2\n\nthe moles of BaCl2 in the solution 0.05 * 2.1 = 0.105\nthe moles of ZnSO4 in the solution 0.0456 * 1.92 = 0.087552\n\nbecause the moles of BaCl2 is greater than that of ZnSO4 so all the moles of ZnSO4 will react completely\n\nthe moles of BaSO4 produced if the percent yield is 100% : 0.087552\nthe grams of BaSO4 is produced with 100% of yield : 0.087552 * 233 = 20.399g\n\nthe percent yield is 16.775 / 20.399 *100% = 82.23%
423379
let me see if i can at least help you with part of your problem, solubility in chemistry is simple, basically the equation will always be:\n\n[Solubility equalibrium constant] = concentration of product divided by concentration of reagents (concentration of each reagents multiplying by each other).\n\nthe easiest way to understand this is to simplify things for you so as you know, a chemical equation always go like this:\n\nreagents <==> products\n\nJust know this from the above, the dynamic here is that when you have an excess of either the reagents or the product, the balance of the equation tend to shift in the opposite direction (ie excess concentration of product will cause more reagents to form and vice versa), in a chemical equation such as the one from above, the reaction can actually go either way, products or reagents can form.  What really matters is the equalibrium constant that I talked about in the first equation. This constant represent the concentration of the products and the reagents when they are in equailibrium or when they are balance, meaning that the rate at which products or reagent form stays the same and there are no change in the concentration of either the products or the reagents so the whole thing is static at the point.   Think about this, more concentration of products wouldn't form unless you have extra reagents to spare that can shift the balance.  When something is saturated it means that it can longer dissolve in a solvent (let's use water to simplify the explanation)so it just accumulates at the bottom, water molecules can only separate so many ions from each other (the reagents) and preventing them from forming a solid product since water itself has limited molecules (if you mix enough table salt in a glass of water, it'll get to the point where it'll accumulate at the bottom and won't dissolve).  If remember correctly, in a chemical equation, when there's different amount of reagents needed to form the product, there's a slight modification to the equation I gave you, for example:\n\n2 CaCO3 + 4 ZnO2 <----> 3 ZnCO3 + 3 CaO2\n\nthen  [equilibrium constant] = [Cao2]^3 x [ZnCO3]^3 divided by [CaCO3]^2 x [ZnO2]^2\n\nnote that you should balance the chemical equations first, I am just giving an example.\n\nhere's a summary of how to compare numbers, if you plug given concentrations of reagents and product into the equation and the number you get is larger than the real equilibrium constant, then the concentration of products is higher than it should be and the equation will shift towards forming more reagents (in the case of the solid product, it'll dissolve) and if there's more reagents, more solid product is gonna form.  Hope that helps. Remember the Chemistry textbook is still your best reference.
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.
1160799
The question is more throroughly stated as "find the minimum whole numbers of each molecule in this reaction so that it balances the number of atoms."  \n\nYou could devise a set of three of algebraic equations so that you wind up with the same amount of oxygen and hydrogens on both sides.  That would be the general way, or you can just do it semi-intuitively with a simple equation like this one.  I will walk you through it. I could just *tell* you the answer, but then you wouldn't learn!\n\n\nSince both reactants are diatomic(*), look at the products:  what is the least abundant atom there?  \n\nNow pick the smallest whole number of the reactant molecule on that still gives you a whole number of product molecules on the product side. \n\nNow see how many of the other molecule you need on the reactant side to get the right number of product molecules (determined in the last step) on the product side. \n\nFor complex reactions with some of the same atoms in multiple reactants and perhaps with multiple products on the other end, it will become harder to just "see" the answer and you will need to rely on the general solution of setting up algebraic equations to solve, so you should learn this technique.  \n\nDo this simple one and maybe a few more, and then see if you can devise the general solution technique yourself to solve them again.  Since you know the answer already you have a ready "reality check" to verify your technique.
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
93462
I hope this helps:\n\nhttp://chemistryformulas.com/
948758
1)write out the equation\n2)find the no of mol of both reactants\n3)compare the stoichiometry ratio of reactant from 1) to find the limiting reagent\n4)compare the mol ratio of the limiting reagent with NH3 to find the no of mol of NH3 produced\n5)theoretical yield=17(no of mol of NH3)
264121
The theory can be used to calculate the mass of product that is expected to be produced when more than one reactant is involved in its formation.\nConsider the reaction below:\n\n2Na2S + Na2CO3 + 4SO2 ---> 3Na2S2O3 + CO2 \n\nGiven the mass of the reactants involved, we would use just one of them as a reference to get the mass of the product, but which reactant should we choose? \n\nHere we find the moles of each of the 3 reactants (given the mass)  and then compare the moles ratio of each in turn with the product and hence calculate the moles of the product Na2S2O3. A limiting reactant is one which gives the least moles of the product. So we see that from which reactant we get the least moles of Na2S2O3, and then with the moles, we can finally calculate the mass of product we expect to get when the reaction is carried out.\nWe choose the limiting reactant for finding out the yield because after any one of the reactants is used up,  the reaction won't proceed further, so this is to calculate the maximum amount of product we can get.
506202
Write out the balanced reaction which occurs.\n\nSince both NaOH and H2SO4 are strong electrolytes (strong base, strong acid), we can assume both dissassociate completely and the reaction produced Sodium Sulfate and water,\n\n2NaOH + H2SO4 --> 2H2O + Na2SO4\n\nFind limiting reactant in order to find which  will be left over (either NaOH or H2SO4).\nIn order to do this, you need to know both the concentrations of the NaOH and the H2SO4 in order to find the initial number of moles of each.\n\nOnce you know how many moles of either OH- or H+ are left over, you can then find the pH, or pOH like this,\n\npH = -log ([H+])\npOH = -log ([OH-])\n\npH + pOH = 14\n\n*Remember to take into account that the final volume of solution is the sum of the intial volumes of each othe reactant solutions when calculating pH.\n\nEDIT:\nYou know, it just occured to me, I dont know why I even brought up pH, that was not what the question was asking.  but everything I said still stands in order to calculate the molarity...you just dont have to go the one extra step and calculate the pH by taking the negative log of the molarity concentration of H+ ions.\n\nAlso, for all of you out there who are just given answers with not other explanation, it would be nice if you stated what the concentration was of....salt? NaOH?  H2SO4?\nBecause of this ommision, I would highly inclined to disreguard their answers as numerically incorrect as well.
834556
the percent yield of 91.34% means: in theory, if you calculate that you will have 100 after the reaction. However, in practical, you just have 91.34\nNow, we solve your problem\n\n4NH3 + 5O2 --> 4NO + 6H2O\n\nit is a balanced equation\n\nI call the moles of O2 as n(02), the moles of NO as n(NO)\n\nn(O2) = 64/32= 2\n\n=> n(NO)= 4/5 n(O2)= 1.6\n\n=> the weight of NO in theory is 1.6*30= 48\n\n=> the weight of NO in practical is 48* 91.34%= 43.8432
1132105
% yield = Actual (5.53g) divided by Theoretical (use periodic table)\n                                times 100%.\n\nFind limiting first by finding moles of H2 and Cl2.  \nH2 moles = 5.95\nCl2 moles= 0.17\nsince H2 to Cl2 is 1:1 in the chemical formula(1H2+1Cl2-->2HCL)\nyou can't react all of the H2 present.  THerefore Cl2 is the limiting reactant.  \n\nYou therefore use Cl2 reactant info to find the Total products formed, because when all Cl2 is reacted the reaction stops.  \nSimply convert moles Cl2 all the way to grams HCL.\nYou get::  12.4g HCl\n\nThis is expected but you actually got 5.53, so the percent yield (use first formula) =44.6%\nI'm checking my work to make sure
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
814637
The equation is  2H2 + CO--->CH3OH\nThe first step in a problem like this is to determine if you have enough of each reactant...or is there a limiting reactant.\n\nYou started with...\n1.73g H2 (MW 2.016) = 0.858mols...and\n16.05g CO (MW28.01) =0.573mols\nBut since 0.848mol H2 would react with only 0.429mols CO you have an excess of CO...too much CO, so your limiting reactant is H2.\nYou get one mol of CH3OH (MW 32.04) for each half mol of H2 per equation, so the theoretical yield is...\n0.429 x 32.04 = 13.7g of CH3OH\n\nCheck my figures for accuracy...which will also give you good practice. :-))\n\nEDITED LATER: If you check this work carefully, you will note that the answer is correct, but the statement is not...it should read:\nYou get one-half mol of CH3OH for each mol of H2 per the equation, so the theoretical yield is...\n1/2 x 32.04 x 0.858 = 13.7g CH3OH\n(Did you find this error?)
149472
Hey I'm studying physical and chemical equilibrium too, currently.The answer is that it will definitely have units when the concentrations have different powers.Just try and work it out with the units and it will be very clear.
947266
In some cases, when the reaction rate is fast, and the reaction goes all the way to completion, one uses the stoichiometric amounts necessary to convert all of reactants.  But most real-world reactions aren't like that.  The reaction rate tends to slow down as product is created.  If one used just the right amounts of reactants to produce the theoretically possible amount of product, one might wait a very long time, or forever: many reactions just reach an equilibrium, and never go to completion.  So, you use a large excess of one of the reactants, which causes all, or almost all, of the other reactant to be converted.