explain how you can determine the amount of a product that will result from a given amount of reactant?
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!
840588
First of all, you need a chemical equation,\n\nNO + O2 ----> NO2\n\nThen, balance it,\n\nNO + 1/2 O2 -----> NO2\n\nNow, what this balanced equation tells you is that the mole ratio between NO and O2 is 2:1. i.e., you need 2 parts of NO to react with every 1 part of O2.\n\nNow, you need to calculate the total number of moles of NO and O2 respectively,\n\n# mole of NO = 17.26/(14+16) = 0.575 mole\n\n# mole of O2 = 13.15/(16+16) = 0.411 mole\n\nAs said before, for every 1 mole of O2, it takes 2 moles of NO to react. From the number of moles of NO and O2, it is clear to see that NO is the limiting reagent. Why? Because if you react all 0.575 mole of NO, it takes only 0.575/2 mole (0.2875 mole) of O2, which you have a lot more than that.\n\nThe limiting reagent is the substance that is NOT in excess, which then determines how much product will be produced (because once this substance is all gone, there will be no reactants to react and give product).\n\nNow, from the balanced equation, you know the mole ratio between NO and NO2 is 1:1. Therefore, the number of moles of NO2 that be formed will be 0.573 moles.\n\nMass of NO2 = # mole * molecular weight\n= 0.573 * (14 + 16 + 16) = 26.465 g
812302
You might just use the formula: \nEntropy of the reaction  = \n(sum of the entropy of products) - (sum of the entropy of reactants)\nIt's as simple as that :)\nThanks.
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..
799022
I'm assuming you really mean "How do you find out how many atoms of a element are in a particular compound?"\n\nJust multiply the subscript by the coefficient\n\nEx. 2 molecules of carbon dioxide. The coefficient is 2 and the subscript for carbon is 1. Therefore there are 2*1=2 atoms of carbon in 2 molecules of carbon dioxide.\n\nIf you just want the amount in one molecule, just substitute 1 for any coefficient you have.\n\nIf you cannot see any subscripts and/or coefficients for the element you are looking at, the subscript and/or coefficient is 1.
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.
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
960715
stoichiometry is \n\nCalculation of the quantities of reactants and products in a chemical reaction. \n\nThe quantitative relationship between reactants and products in a chemical reaction.
988889
understanding the reaction meachanism \neffect of catalysts etc....\ndetermining the amounts needed
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%
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.
832153
2 moles of reactant  (C2H4F2) give you 4 moles of HF\nC2H4F2 : 66 g/mol\nHF: 20g/mol\nTherefore\n79,3g  C2H4F2*(1mol  C2H4F2/66g  C2H4F2)*(4 moles HF/ 2moles C2H4F2)*(20g HF/ 1mol HF)= 48,1g HF\nThe amount of HF you are expected to get with 100% yield is 48,1g. If you got 18g, then\n(18g/48,1g)*100= 37% yield
199900
The easiest way is to simply find literature listing delta H for the reaction.  Otherwise you have to do one of 2 things.  \n\n1. Find the enthalpy of formation for the product of the reaction which should be what you're looking for.\n\n2. Use the equation G=H-TS, plugging in the Gibbs free energy, temperature, and entropy values for the reaction,then solving for H, the enthalpy
815531
It is simply sum of products minus sum of reactants\n\n[H2O + N2] - [O2 + NH3]\n\nSince this is a 1 to 1 ratio reaction, the coefficients are one. (if there was a coefficient for NH3, you must multiply the enthalpy value of NH3 by that # because that is how many moles you have, and the value given is for only one mole.)
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.
861726
You will have to first find the standard enthalpy values of the reactants and products. They're given in the appendices of any standard GenChem or PhysChem books.\n\nThen, you subtract the sum of enthalpies of reactants from the sum of enthalpies of the product to get the enthalpy of the reaction.
600524
Give the mass of one reagent in a reaction, and ask for the mass of another. A word equation or chemical equation must be provided.\n\neg:\n\nWhat mass of nitrogen gas is required to produce 45.0 g of ammonia, according to the reaction below ?\n\n         N2   + 3 H2   ---> 2 NH3\n\n\nTo solve the problem\n\na) convert mass of NH3 to moles (2.65 mol)\nb) use stoichiometry - find moles of N2 needed (1.32 mol) \nc) convert moles of N2 to mass (37.0 g)