For the balanced equation, if the reaction 79.3 g C2H4F2 produces 18 g HF, what is the percentage yield?
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
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.
942373
From the reaction\n\nS8(g) ===> 4S2(g)\n\nKp = (pS2(g))4/pS8(g) where p is the partial pressure\n\nGiven the partial pressure of S8(g) at equilbrium 0.25 atm, clearly 0.75 atm of S8(g) has been consumed to produce (from the equation) 4 volumes of S2(g), Given the volume was fixed this mean dS2(g) = 3 atm\n\nSubstituting into Kp yields\n\nKp =3^4/0.25 = 324\n\nThus Kp at 1325 K is 324
749681
find the stoichiometric yield.  then multiply by .8962 voila.
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
955656
The reaction\n\nF2 + O2 ===> 2F2O\n\ndH = 2 dH F2O - (dH F2 + dH O2) wher d is the delta symbol and using the heats of formation\n\nSince F2 and O2 are in the elemental state, the heat of formation is 0 (by definition)\n\ndH = 2 x 24.5 = 49 kJ/mol\n\nDelta H for the above is 49 kJ/mol\n\nI got the heat of formation for F2O (difluorine monoxide) from Wikipedia.
514226
I think this is how it goes...\n\n8F+3z(4h-2h)+(h+3)...\n\n8f+12zh-6Zh+h+3...\n\n8f+6zh+H+3
329382
c2h4+ 3o2 -> 2co2 + 2h2o is balanced\n\nBefore, there are 2 carbon, 2 hydrogen and 6 oxygen\nAfter is the exact same.   \n\nIs there something else you are asking?
655492
NaHCO3 + HCl ----> NaCl + H2CO3\n\n10 grams\n(10/84) = .12 moles\n\nAs the yield is 75%\nNo. of moles of NaCl are (.12*.75) .09\n\nWeight of the NaCl is (.09*58.5) 5.265 grams
412413
I'm assuming you're asking for the reactions.  Here are a couple:\nH2 + F2 → 2HF \n\n4Fe + 3O2 → 2 Fe2O3
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
192134
1.74g should = 17.4g approx.\n\nso,\n\n x  = H (Hydrogen) = 1.008g
532149
I don't know if this is right ... but check it and tell me\n\nV*C KOH = V*C HBr\n\nV KOH = 18.6ml C KOH = 0.752M\nV HBr = 22.1ml C HBr = ?\n\n18.6*0.752 = 22.1*C HBr\n\nC HBr = 0.6329M\nIts seems reasonable cause the difference between concentrations isn't significant
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
780875
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O