how do i find the theoretical yield of a chemical equation?
464799
I assume that you have an equation such as A and B react together to give C. Work out the molar ratios of each, i.e. is it 1 equivalent of A reacted with 1 equivalent of B to give 1 equivalent of C?\nAssuming it is a 1:1:1 ratio, then look at the amounts you actually reacted - chances are that you did not use 1 mole of each of the reactants. Work out the molar quantities of each reactant. (mass used/molecular weight of compound)\n\nIf you used 0.5 moles of each, then you would get 0.5moles of product and just convert this to a mass.\n\nIf one of the reactants is in excess, e.g. 0.5 moles of A and 0.8moles of B was used, ignore B because as there is only 0.5 moles of A you would only make a maximum of 0.5 moles of C.\n\nHope 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%
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
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!
739541
CH4 + 2O2-> CO2 + 2H2O
362155
Any case, the best way to calculate the yield of the product is from moles ratio.\nJust check howmany moles of reactants are participating and how many moles of product should form in theory... according to that you can get as idea about the yield of the product.\n\nif you can provide the reaction, i will help you more in detail\n\nGood luck!
781593
I always have to derive the equations from scratch:\n\nKa = [H+][A-]/[HA]\n\nlog(Ka) = log(H+) + log(A-) - log(HA), multiple both sides by -1\n\npKa = pH + log(HA) - log(A)\n\npH = pKa + log(A-/HA)\n\n4.54 = 4.74 + log(A-/HA)\n\n-0.20 = log(A-/HA)\n\n10^(-0.20) = A-/HA = 0.631\n\nplease check for errors
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
290031
Balancing chemical equations is easy, but it's a bit of a trial and error process. Basically, you've got the the arrow in the middle, so the number of each element on one side of the arrow has to equal the number of that particluar element on the other side of the arrow. So start off by balancing any unique elements that only appear once on a particular side. Then balance your nitrogens, oxygens and carbons in that order. The last thing that should be balanced is your hydrogens. I can go into further detail if you need me to, I just don't know how much you know. \n\nFor your other question and yield is how much of something you get from a reaction. For instance, if I mix 100 moles of one substance and 100 moles of another substance, I can calculate a theoretical and experimental yield. For instance, I can say that I expect all 200 moles to react and form a single product. Since i expect it all to react, my THEORETICAL yield is 200 moles or 100%. Say I actually run the experiment and I discover that only 100 moles of my intended product formed, then my EXPERIMENTAL yield is 100 moles or 50%.
379843
Using the formula for pH: pH = -log(H+)\nwe can find the molarity\n\nFirst, we plug in what we know\nThe pH is 2.65\nand the H+ is the concentration of hydrogen ions measured in moles/ liter. Hey, that's what we're looking for!\n\n2.65 = -log(moles/liter)\n\nusing algebra we can solve for moles/liter\n\n-2.65 = log(moles/liter)\n\nLet's raise 10 to the power of each side\n\n10^(-2.65) = 10 raised to the negative 2.65 power\n10^(log[moles/liter]) = 10 raised to the power of log(...)\n\nOk, so..\n\n10^(-2.65) =10^(log[moles/liter])\n\nNow 10 with log as a power cancel each other out so...\n\n10^(-2.65) = moles/liter\n\nmoles/liter is the molarity, let's call it "M"\n\nM = 10^(-2.65) \n\nM = approx. 3.16 x 10^(-3)
919460
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.
950743
theoretical amount of product=10g lead acetate/ mw lead acetate x mw lead iodide\n\ntake the result of this formula as the denominator and divide 5.00 grams product/theoretical x 100%
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
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
932271
yields "H2 + Zn2O3"\n\n^^!
937526
If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)\nFill in the atomic masses of H = 1.00797 and N = 14.0067:\n[3 x 2 x (1.00797)] +[ 2 x (14.0067)]  = 2 x (14.0067) x 3 x (1.00797) = 34.06122\nThis is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:\n2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.
84780
!!!!!!!!!!!!!!!! \nI think rather u wanted to ask how can u predict the formula of a chemical reaction!!\nFirstly, we have to determine the moles of reactant taking part.(It can be done easily, think A reacts with B, just mix 1 mole of A with the same amount of B.)\nThen determine what are the different products and their amounts.\nAll u have to do is some calculations..................\nRemember its not as easy as it sounds.