What is the theoretical yield of   CH3OH ?   I.E., how many grams of   CH3OH   should be formed?
963470
CO + 2H2  --->  CH3OH\n\nmoles H2: 2.00g/ 1g/mol = 2.00 mol\nmoles CO: 18.50g / 28g/mol = 0.6607mol\n\nyour limiting reactant is CO since for every 0.6607 moles of CO, you'll only need 1.321 moles H2. there's an excess of 0.6790 moles H2.\n\ntheoretical yield, therefore, is (1.321/2)*100 =  66.05%\n\nbased from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,\n\nweight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams
904652
CH3CH2C(CH3)3.\nI guess you forgot to type the C before the parenthesis. If you didn't forget, then none of the answers is correct (you would be missing a carbon atom).\n\nChem-man I would have contacted you in private about this but you do not give us the option to e-mail you. I have seen several of your answers prompting askers to verify your answer through echemistryhelp. I checked the website and it is not a free one; you post a question and pay for the answer. I'd say that this is a form of indirect advertisement  which is against the community guidelines. Please refrain from doing that or I will have to report you.
804090
ooo this will be fun!\n\nok..\nA)\n2.207g of H2O MW. 18.0\nmol = weight of sample (g) / molar weight (g/mol) (n=m/M)\nn = 2.207 / 18\nn = 0.1226 moles of H2O\n\nnow, since there is 2 hydrogens in H2O, then when the molecule is broken down, there is twice as many moles of hydrogen then the overall molecule,\ntherefore,\n\n2x 0.1226 = 0.2452 moles of H.\n\ndetermining number of moles of CO2 is also helpful,\n4.313g, n = 4.313 / (12.01 + 16 +16) = 0.0980moles\napprox, 5(moles H2O) :  4(moles CO2)\n\nB) \nworking the number of moles of this compound will be useful:\nn = 1.816 / 74.122 = 0.0245 moles, which is 1/10 of the moles of H produced. (or 1:5 of the moles produced of water)\n\nif you think about it, the reaction is something like this:\nCH2 + O2 --> CO2 + H2O\nclearly the above doesn't make sense, because 1) it isn't balanced, and 2) the reactant needs a higher molecular wt.. now thinking about it, the following can be deteremined:\nand from the ratio of moles found above:\n       1     : (excess):  4        :    5\nC4H10 + 13/2O2 --> 4CO2 + 5H2O\n\nnow, I amoung most allow fractions in equations, but if you don't, multiply through by 2.\n2C4H10 + 13O2 --> 8CO2 + 10H2O\n\nnow, you want the empirical forumla,\ntherefore,\ndivide by highest common factor,\nC2H5 \n\nthe empircal fomula is just a ratio and doesn't make huge sense, as the formula for an alkane is CnH2n+2\nhope that makes sense :)
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!
702103
1) Determine how many moles of C3H8 we have:\n147 g/ (3*12.01+8*1.008) =3.33 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nC3H8 + 5 O2 -> 3 CO2 + 4 H2O \nWe need 5*3.33 = 16.67 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. Determine at the given conditions:\nV(1mol) = 22.4 * (303/273) = 24.86 liter \n\nTherefore 16.67 mol = 414 liter = 0.414 m^3
805524
Hydrogen and nitrogen react to form ammonia in what is called the Haber process.\n\nThe reaction is as follows:\nN2 + 3 H2 <--> 2 NH3\n\nThis reaction will give about a 85% yield at 200°C (473 °K) and 200 atm of pressure.  This is typically what is done in industry.\n\nWith special considerations to safety, you can get about 100% yield at 200°C and 750 atm of pressure.\n\nThe only remaining question is, how much do you want to make?\nI will assume 100 kg per batch.  You can recalculate for different amounts.\n\nThe molecular weight of ammonia is 17 g/mol, so 100 kg will have 100 kg / 0.017 kg/mol = 5882 mol\n\nTo produce 5882 mol of NH3, you need 2941 mol of N2 and 17646 mol of H2\nSo, the maximum number of moles of gas will be at the beginning of the reaction: 2941 + 17646 mol = 20587 mol\n\nUsing the ideal gas law:\nPV = nRT, where R is the gas constant and n is the number of moles\nV = nRT/P = 20587 mol * .082057 L*atm / mol*K * 473°K / 750 atm = 1065 L = 1.065 m3
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.
555664
The formula could be (C6H12OH)n, in which case it could be synthesised from the corresponding aldehyde. Expensiveness depends on how much and how pure of a product you want.
250273
Enough to cancel out the H's. For the reaction with 2H's, add 2OH's to both sides. For the reaction with 3H's, ad 3OH's to both sides.\n\nH + OH --> H2O\n\nEDIT: Oh, well in that case, the H's cancel out, so you have just one H on the side that previously had 3. Then, just add one OH to both sides.
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
777003
CH3COOH + NaOH ---> NaCH3COO + H2O
750464
My chemistry is pretty rusty, but let's see.\n\nI'll assume that you have an infinite amt of O2, and that you have 68.12 grams of NH3.\n\nThe equation isn't balanced, so you need to do that first (you have 3 hydrogens on one side, but two on the other).  Balance to get:\n\n2(NH3) + (2.5)(O2) --> 2NO + 3(H20)\n\n4(NH3) + 5(O2) --> 4NO + 6(H20)\n\nNow, given that you have 68.12 grams of NH3, you need to convert this into moles of NH3, as Stoichiometry deals with molar ratios.  To get the conversion ratio, look at the periodic table.  Under each element, there should be an atomic mass.  For oxygen is 16.something, Hydrogen 1.008 ( i think ).  Find it for N and H.  (this is where I get hazy on the math).  I believe: take 3 * atomic mass of hydrogen + atomic mass of nitrogen, let's call it M  (it should be in grams/mole)  So to get moles, multiply 68.12 * (1/M).  \n\nNow, our equation above shows that for every 4 moles of NH3, we get 6 of H20.  # moles NH3 * (6/4) = # moles of H20.  Now, do the reverse operation of H20 to get how many grams is produced.  H20 has an atomic mass of something like 18.016.  So, # moles H20 * 18.016 = # grams H20\n\n\nI hope that's somewhat correct, and if not, leads you in the right direction at least
800696
19) The number of moles of CO in 86g is: 86 / 28 = 3.07 moles.\n\nCO + 2H2 ----> CH3OH\n\nFrom this equation, 1 mole CO, when reacting with H2, is supposed to produce 1 mole CH3OH.\n\nTherefore, 3.07 moles of CO should produce 3.07 moles of CH3OH, which equals 3.07 * (12 + 4 + 16) = 98.29g.\n\nIn fact, we've got only 68g, hence, the percent yield of CH3OH is:\n(68 / 98.29) * 100% = 69.19%\n\nThe next question is essentially the same.
145111
The reaction formula is\n\nCH2COOH                           \nCH(OH)COOH (aq) + NaHCO3 (aq)----> \nCH2COOH                            \n\nCitric Acid\n\n       CH2COONa\n CH(OH)COONa    (aq)+ H20 (l) + CO2(g)\n       CH2COONa
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
428107
first count molecules of carbon in CH3Co2H, in this case its one, 12.01 grams of carbon in each mole of CH3Co2H, divide 12.01 by the total weight mass of CH3Co2H which is 133.91 and you get .0897 or 8.97%
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%