What is the theoretical yield of   CH3OH ?
814637
The equation is  2H2 + CO--->CH3OH\nThe first step in a problem like this is to determine if you have enough of each reactant...or is there a limiting reactant.\n\nYou started with...\n1.73g H2 (MW 2.016) = 0.858mols...and\n16.05g CO (MW28.01) =0.573mols\nBut since 0.848mol H2 would react with only 0.429mols CO you have an excess of CO...too much CO, so your limiting reactant is H2.\nYou get one mol of CH3OH (MW 32.04) for each half mol of H2 per equation, so the theoretical yield is...\n0.429 x 32.04 = 13.7g of CH3OH\n\nCheck my figures for accuracy...which will also give you good practice. :-))\n\nEDITED LATER: If you check this work carefully, you will note that the answer is correct, but the statement is not...it should read:\nYou get one-half mol of CH3OH for each mol of H2 per the equation, so the theoretical yield is...\n1/2 x 32.04 x 0.858 = 13.7g CH3OH\n(Did you find this error?)
818293
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
384774
CH3CH2OH\n\nC ~ 12\nO ~ 16\nH ~ 1\n\n2*C + 1*O + 6 * H = 2*12 + 1*16 + 6*1 = 46 g/mol
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
948758
1)write out the equation\n2)find the no of mol of both reactants\n3)compare the stoichiometry ratio of reactant from 1) to find the limiting reagent\n4)compare the mol ratio of the limiting reagent with NH3 to find the no of mol of NH3 produced\n5)theoretical yield=17(no of mol of NH3)
132050
oh shit.ok\nn2h2+(ch3)2n2h2+3n2o4 = 5n2 + 6co2 + 5h2
807508
Most of the info provided is superfluous.\nSimply put, you burned something and got 4.161g H2O (MW=18.016), all H coming from the sample.\nSo you formed 4.161/18.016 = 0.23096 mols H2O, and since each mol H2O contains 2 mols H, your sample contained 0.4619 mols of H.\n\nBTW, it is easy to deduce the empirical formula of the compound:\nNo hydrocarbon can have a MW of 60.095, and since only CO2 and H2O were formed, the compound  must contain O...this works out to be C3H8O, which means it is probably propanol (n or iso) or methyl ethyl ether.
592234
3g of hidrogen (plus 14 g of nitrogen) make up 17g of ammonia (H+H+H+N=1+1+1+14), so:\n\n3g  H____17g NH3\n50g H____=283.33g NH3\n\nI assumed a 100% efficient reaction (that means every bit of reactant will actually react).
702782
CH2OHCl: 66,45g/mol\nDo you mean 4dm^3 or a cube with 4dm length  edges?\nIf you mean 4dm^3 then\n4dm^3= 4*10^3 cm^3 = 4*10^3 mL= 4L\n2M= ?/4L\n?= 8 moles of CH2OHCl \nif you meant 4dm edges then\n4dm= 4*10^1cm\n(4*10^1cm)^3= 6*10^4cm^3 (actually it is 6,4 but because of the lack of figures you must report 6 instead of 6,4)\n6*10^4cm^3= 6*10^4 mL = 6*10^1 L\nTherefore\n2M= ?/6*10^1L\n?= 1,2*10^2 moles of CH2OHCl \n\nPart II\nCH4 + 2O2 → CO2 + 2H2O\n7dm of O (again: 7dm edges or 7dm^3???)\nfor 7dm sides\n(7*10^1cm)^3= 3*10^2 L \nO density = 1,43 g/L\nTherefore: 5*10^2g or 0,5 kg of Oxygen\n4g CH4*(1mol CH4/16g CH4)*(1mol CO2/1mol CH4)= 1*10^1g of CO2\nIf you compare the amount of Oxygen and methane and the molar ratios there is no need to do the same for O. Obviously CH4 is the limiting reagent\nCO2 density: 1,98*10^-3 g/mL\n1,98kg/m^3\nIf 5dm (again the same problem) were produced, then\nfor 5dm edges: 2*10^2g of CO2 (obviously not this one)\nfor 5dm^3: 5*10^3 cm^3 of CO2: 1*10^1g of CO2 this is it!)\nNow the yield\nIf you would have added more figures we would have had\n(9,9g/11g)*100= 90% yield\nBut because of the lack of figures we end up with\n(1*10^1g /1*10^1g)*100= 1*10^2 % (you got the maximum yield)
749482
1-C Hydroboration puts -OH on least substituted C; anti-Markownikov\n2-B Correct\n3-A Correct\n4-C Br goes on the center C; product has two identical CH3CH2- groups on center C, therefore symmetrical and achiral\n5-A Correct\n6-A Correct\n7-B Correct\n8-A Correct\n9-C Markownikov addition puts H on C next to C-CH3 and OH onto HO-C-CH3
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!
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 :)
868179
2Rb + 2 H20 --> 2RbOH + H2
764178
Trick question, multiple answers. . .\n\n1. CH3(CH2)3OH\n2. C2H5CH(OH)CH3\n3. (CH3)2CHCH2OH\n4. (CH3)3COH\n5. (C2H5)2O\n6. CH3(CH2)2OCH3\n7. (CH3)2CHOCH3\n\nOk, #4 is tert-butanol.  All of these are structures for C4H10O\n\ngo to http://www.cem.msu.edu/~reusch/VirtualText/intro3.htm and scroll down to the isomers section.  All the diagrams are listed there.
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.
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
955014
First you have to know the chemical equation, which is 3H2 (g) + N2 (g) <=> 2NH3 (g)\n\n\nThen, you calculate the number of moles for each reactant that is available, using the mass given.\nno of moles of H2\n= mass of H2 / relative molecular mass of H2\n= 2.55g / 2 x1                       ---> there are 2 H atoms in H2\n= 1.275\n\nno of moles of N2\n= mass of N2 / relative molecular mass of N2\n= 13.65 / 2 x 14\n= 0.4875\n\n\nThen you need to find which is the limiting reactant. In this case, the limiting reactant is H2. The reaction will stop due to the lack of H2 to react with N2, even though there is excess N2.\n\nUsing the no of moles of H2,\nno of moles of NH3\n= (1.275 / 3) x 2               ----> make use of the mole ratio\n= 0.85\n\nThus, calculating the mass of NH3,\nmass of NH3\n= no of moles x relative molecular mass\n= 0.85 x (14 + 3 x 1)\n= 0.85 x 17\n= 14.45g\n\nBy the way, it is called theoretical yield because it is highly impossible that the amount you collect at the end of the experiment is the exact amount you have calculated. :D\n\nCheerios.