What is the theoretical yield of   CH3OH ?   I.E., how many grams of   CH3OH   should be formed?
969093
H2 = 2g/mol   2.05g = 1.025mol\nCO = 28g/mol    15.85g = .57mol\n\n2H2 + CO --> CH3OH\n\nmust be reduced to \nH2 + (1/2)CO --> (1/2)CH3OH\n\n1.025mol + (1/2).57mol --> the quantity of the limiting reactant which is .285mol CH3OH which at 32g/mol = 9.12g
474531
CH3COOH + NAOH =.CH3COONA +H2O
900980
A 1,2-diol would form.\n\nEthene would form ethan-1,2-diol, (CH2OH)2.
502594
If your compound has the formula C6H10, then the molecular weight would be 82 grams/mole\n\nYou could have one of 13 different compounds -- \n2,3-dimethyl-1,3-butadiene\n3-3-dimethylbutyne\nCyclohexene\nany one of the five hexadienes\nany one of three hexynes\n4-methyl-1-pentyne\n4-methyl-2-pentyne\n\nWhat is the rest of the question regarding the 5.3*10^-23 odr?
776525
Since you are pretty, here's how:\n\n1. Compute for the number of moles NaOH needed to prepare the solution.\nmole NaOH: 1 L * 0.5 mol/L = 0.5 mol\n\n2. Get the weight NaOH you need by multiplying the number of moles to the molecular mass of NaOH.\nweight NaOH: 0.5 mol * 40g/mol = 20g\n\n3. To prepare, weigh 20 grams of NaOH pellets. Dilute this with water to 1000 mL in a volumetric flask.
797059
118 gram of C3H8 is 118/(3*12+8) = 2.682 mol.\nThe reaction formula is C3H8 + 5 O2 -> 3 CO2 + 4 H2O, therefore 5*2.682 = 13.4 mol O2 is needed.\nAt STP this is 13.4 * 22.4 = 300 liter.\nSTP corresponds to 273 K and 101.3 kPa (760 torr).\nAs PV/T is constant, we can write\nV = (P_STP/P) (T/T_STP) V_STP = (100/101.3) (302/273) 300 =328 liter = 0.328 m^3.
166454
I can supply you said chemicals.\n(C2H5NH)(CH3)C6H3OH \nPlease contact me by email: halsliu@gmail.com
543827
If you know the grams, which you do on one...its easy.  To find the grams from ml, you need to know the density.  Density is in grams/ml, so if the density was 2 g/ml and you had 4.5 ml, then you would have 9 grams.  So you take the grams and you need to find moles to calculate percent yield.  To do this, find the total molecular weight (for instance, water is 18 b/c of two H's which are 1 and one O which is 16).  When you have the grams and molecular weight, divide the grams by the molecular weight.  This will give you the moles, since the molecular weight is grams per mol.  Do this for every reagent or thing you use.  Whicever one has the lowest amount of moles is your limiting reagent because once that is gone, the reaction stops!  Now the theoretical yield is found by doing stoiciometry.  So whatever your limiting reagent is...that number is your theoretical yield for this problem.  Then weight the product when you are done, and find the amount of moles you have by oncve again dividing how many grams you have by the molecular weight.  Then divide that by your Theoretical yield and that is your percent yield!  Looks harder than it is.
814744
In the presence of the proper catalyst:\n\nCO + 2 H2  -->  CH3OH\n\n1.40 g of H2 = 1.40 g / 2.016 g/mol = 0.694 moles H2\n\n13.85 g CO = 13.85 g / 28.01 g/mol = 0.4945 moles of CO\n\nIt takes 2 mole of H2 for every 1 mole of CO, so only 0.347 moles of CO will react.  This will give 0.347 moles of CH3OH:\n\n32.04 g/mol * 0.347 moles = 11.12 grams\n\n11.12 is your answer.
655398
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.
900861
As usual, we first check for a balanced equation.  It isn't, so we fix it: 2 LiOH + CO2 -> Li2CO3 + H2O.  Also as usual, get the molecular weights: LiOH = 24, CO2 = 44, and the others don't matter.  There are 4 moles of CO2, so we need 8 moles of LiOH, and that is 192 grams.  Extra credit: find the error in the next response.
700367
1) Balance the reaction equation:\nPb(NO3)2  + 2 LiBr -> PbBr2 + 2LiNO3\n\n2) Calculate the amounts we start with\nPb(NO3)2 :  2.05 L * 0.0550 mol/L = 0.1128 mol\nLiBr : 1.52 L * 0.0618 mol/L = 0.0939 mol\nAs we need 2 moles of the latter for 1 mole of the former, the reaction is limited by the amount of LiBr\n\n3) Calculate the amount of PbBr2 produced\n0.0939 mol of LiBr will react with 0.470 mol of Pb(NO3)2 to produce 0.0470 mol of PbBr2. This is 0.0470 * (207.2 + 2* 79.9) =17.249 gram\n\n4) Calculate the yield\nThe yield is 15.784 / 17.249 = 91.51%
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 :)
777003
CH3COOH + NaOH ---> NaCH3COO + H2O
1121888
Well you got C3H3Br + CH3S,  I'd guess something like ethane and bromosulfide.. C2H6 + BrS..
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.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)