suppose that exactly 24.9 g of each reactant is taken. Determine which reactant is limiting, and calculate wha
445261
Try it yourself, its not that bad. \nHere is how:\n1) Set up a stoichometry ratio.\n2) Using the mass that they give you, calculate the theoretical yield of each product.\n3) Whichever product has a smaller theoretical yield, analyze the reactants which caused it to be such.\n4) The one that 'limits' the amount of products is the limiting reagent.\n5) Using that value, calcuate the amount of the other reactant that is necessary. Subtract from the orginal value to find the excess reactant.
973272
ok, first go to moles.\n\n24.0g LiOH (1mol/23.948g) = 1.0022mol\n\nnow get liters using molarity\n\n1.0022mol (1L/0.153mol) = 6.55L or 6,550mL \n\nRight, there ya go, remember that you have 3 sig.figs. so you may want to represent significance by using scientific notation. So...\n6.55x10^3mL\n\nIf you're ever stuck, just simply go to moles and that will get you started (and usually gets you some points on tests).
877279
First you need to determine your limiting reagent in the first step.\n\nMolecular mass of NH3 = 17 g/mol, O2 = 32 g/mol\nMol of NH3 = 5,700,000 g/17 g/mol\nMol of NH3 = 3.35E5 mol\n\nMol of O2 = 57,000,000 g/32 g/mol\nMol of O2 = 1.78E6 mol\n\nSo based on the first equation (1), NH3 is the limiting reagent.\nFor every 4 mol of NH3, one gets 4 mol of NO.  So the first reaction produced 3.35E5 mol of NO.\n\nThe second reaction produces 2 mol of NO2 for every 2 mol of NO used, so the second reaction produced 3.35E5 mol of NO2.\n\nThe third reaction produces 2 mol of HNO3 for every 3 mol of NO2 used, so this final reaction will produce 2/3 x 3.35E5 mol of HNO3 = 2.23E5 mol.\n\nThen use molecular mass of HNO3 to calculate how much of it (mass) that you made.\nHNO3 molecular mass = 63 g/mol\nmass = mol x molar mass\nmass = 2.23E5 x 63 g/mol\nmass = 1.41E7 g of HNO3\n\nOf course this assumes 100% yields in all the above noted reactions.
986414
Start by balancing the equation:\n2Ca3(PO4)2 (s) + 6H2SO4 (aq)-----> 6CaSO4 (s) + 4H3PO4(aq)\n\nThen figure out the molar masses of each of the reactants:\nCa3(PO4)2 = 3(40.078) + 2(30.973762) + 8(15.9994) = 310.176724\n\nH2SO4 = 2(1.00794) + 32.065 + 4(15.9994) = 98.07848\n\nDivide each of your known masses of reactants by its respective molar mass to determine how many moles of each you have, and divide this number by the coefficient (the number in front) of the reactant to determine how many "parts" of each you have:\n(1000/310.176724)/2 = 1.612 parts 2Ca3(PO4)2\n(980/98.07848)/6 = 1.665 parts 6H2SO4\n\nTake the smaller number (since once you run out of one reactant, the reaction is over) and figure out how many moles of calcium sulfate and phosphoric acid you'll produce:\n(1.612)(6)(40.078 + 32.065 + 4(15.9994)) = 1316.7518832g CaSO4\n\n(1.612)(4)(3(1.00794) + 30.973762 + 4(15.9994)) = 631.872933536g H3PO4\n\nSince the total mass of reactants and products needs to be the same, you can check your (er, my) work by totalling up both sides:\n(1.612)(2)(310.176724) + (1.612)(6)(98.07848) = 1948.625g\n1316.7518832 + 631.872933536 = 1948.625g
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..
727216
use beers law to figure out the concentration and then take that to mole ratio
805333
sorry i do not have a periodic table with me now... but if i never remember wrongly the Ar of Al is 27.0, Ar of Cl is 35.5 and Ar of H is 1.0\n\n1. thus the no. of moles of Al is larger than HCl, thus HCl is the limiting reactant.\n\nno. of moles of Al= 14.6/27.0=0.541mol\nno. of moles of HCl=14.6/36.5=0.4mol.\n\nsince HCl is the limiting reactant, number of moles of AlCl3 and H2 will be 0.4mol also.\n\n2. mass of AlCl3= no. of moles of AlCl3xMr of AlCl3 \n=0.4x133.5=53.4g\n\n3. mass of H2=no. of moles of H2xMr of H2=0.4x2.0=0.8g
387120
You're close.  To find out how much uranium oxide is formed you have to determine how many moles of each starting material you have available and which is the limiting reagent.\n\nU + O2-> UO2\n\nMoles U = grams U/molecular weight = 100g/238.03g/mol\n=0.42moles uranium\n\nMoles O2 = grams O2/molecular weight = 37.2g/32g/mol\n=0.156moles O2\n\nFrom the chemical equation; 1 mole U + 1 mole O2 yeilds 1 mole UO2.  So from the above calculations we can see that the amount of O2 is limiting.  Thus the number of moles of UO2 formed must be equivalent to the #moles O2 (because it's limiting).  You can then calculate the grams of UO2:\n\ng UO2= moles UO2 formed * molecular weight UO2\ng UO2 = 0.156moles * 270.03g/mol\ng UO2 = 42.12g\n\n\nTo fully oxidize 100g of uranium, the number of moles of oxygen gas would have to be equivalent to the # of moles uranium.\n\nFrom the calculations above, we know that 100g U= 0.42 moles\n\nSo the # moles O2 required for complete oxidation is 0.42 moles.  Therefore the number of grams O2 can be calculated by using the molecular weight of O2:\n\ngrams O2= #moles O2* molecular weight O2\ngrams O2= 0.42* 32g/mol= 13.44grams O2 required
904268
just add 3  zeros or multiply by 1000 (which is the same as adding 2 zeros o_O)\n\nexample: 78g=78000mg
649652
If oxygen is your limiting reactant, then the amount of CO2 produced is dependent solely on the amount of oxygen.\n\nThe balanced equation being 2CO + O2 ---> 2CO2, you will generate 2 moles of carbon dioxide for every mole of oxygen, so the answer is 100 moles.
628545
11.  Balanced equation  2C + 2H2O ---> CH4 + CO2\n\n1.26g / 12 g/mol = 0.105 moles C = 0.0525 moles CH4 = 0.84 g\n\n12.  a.  P4O10 + 6H2O ---> 4H3PO4\n     b.  100 g P4O10 / 284 g/mol = 0.352 mol P4O10\n         200 g H2O / 18 g/mol = 11.1 mol H2O; P4O10 is the limiting reactant, so theoretical yield of H3PO4 is 4 * 0.352 * 98 g/mol = 138.0 g\n     c.  % yield = 126.2 / 138.0 x 100% = 91.4%\n\n13.  Actual yield cannot exceed theoretical yield, as you can't create mass.  If it does come out higher, then there are most likely impurities present, or unreacted reactants.
485893
Check on the internet for the ratios used by comercial factories, then prepare a solution from glycol and water in the given proportion. That's all!!!
192134
1.74g should = 17.4g approx.\n\nso,\n\n x  = H (Hydrogen) = 1.008g
262778
73.93 grams\nFirst you use the Ideal Gas Law: PV=nRT >P is pressure in atm., V is volume, n is moles, R is the constant: .0821, T is the temperature in Kelvins> (1atm.)(3.16L)=(n)(.0821)(305K) > .126195563 moles. Now you use a proportion. 9.33grams/.126195563moles = x grams/ 1 mole > 73.93 grams
867375
=19g-4h+4-20g+20\n\n= -g-4h+24
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
973794
moles = concentration* volume\n\nfind the moles of the substance with given concentration and volume\n\n\nwrite and balance the equation. determine the moles equivalence of every reactant then use the equation above