10.The reaction of 10.0 g H2 with 10.0 g O2 yields 7.81 g H2O. What is the percent yield of this reaction?
823484
First, make a balanced equation:\n\n2H2 + 02 → 2H2O\n\nNow find the theoretical yield (use O2 because it's the limiting reagent):\n\n10.0 g O2 * (1 mol O2/31.9988 g O2) = 0.313 mol O2\n\n0.313 mol O2 * (2 mol H2O/1 mol O2) = 0.626 mol H2O\n\n0.626 mol H2O * (18.009 g/1 mol H2O) = 11.3 g H2O (this is the theoretical yield)\n\nPercent yield = (actual yield/theoretical yield) * 100\n\nPercent yield = (7.81/11.3) * 100\n\nPercent yield = 69.1%\n\n\n* Please vote my answer as the best since I went into greater detail.  Thanks.
929547
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
605005
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
739541
CH4 + 2O2-> CO2 + 2H2O
411083
If the pH is 4, the pOH is 10 (pH + pOH = 14).  10^(-pOH) = [OH-].\n\nAnswer is 1E-10
977270
first you need a balanced equation.\n\n2Mg+O2===>2MgO\n\nNext convert everything given to mols.\n\n10.000 g Mg X 1 mol Mg / 24.30 g Mg (NOTE: this number is from the periodic table of the Elements) = 0.4115 mol Mg\n\n5.267 g O2 X 1 mol O2/ 32.00 g O2 (this is the molar mass of O2)= 0.1646 mol O2\n\nAt this point pick one of the mols you calculated.\n\n0.1646 mol O2 X 2 mol Mg (from balanced equation) / 1 mol O2 (from balanced equation)(for every 2 mol of Mg there is 1 mol of O2)= 0.3292 mol Mg needed to react with all of O2\n\nSince more Mg is available then needed (0.4115 mol > 0.3292 mol) this implies that O2 is the limiting factor. \n\nUse the limiting factor (O2) to get the mass of the MgO\n\n0.1646 mol O2 X 2 mol MgO / 1 mol O2= 0.3292 mol MgO\n\n0.3292 mol MgO X 40.30 g MgO/ 1 mol MgO = 13.27 g MgO\n\nessentially for all the other ones you are doing the same thing as before but with different numbers.
885936
we have PV = m/M * R * T\nT = t(celsius) + 273\nR = 0.082 = const\nM is the molecular mass\nP is the pressure (1atm = 760 mmHg)\n\nso M = (m*R*T)/(P*V) = 3.6 * 0.082 * (23.4+273)/(4042.4/760 * 0.183 )= 90 g/mol\n\nwith the percentages we can find the compound \n\nthe formula is CxHyOz (x,y,z is the number of atoms)\n\nx : y : z = %C/M(C) : %H/M(H) : %O/M(O) = 40/12 : 6.71/1 : 53.28/16 = 3.33 : 6.71 : 3.33 = x : 2x : x\n\nwe can use this table to find the answer\nx=1 ; M = 30 : CH2O\nx=2 ; M = 60 : C2H4O2\nx=3 ; M = 90 : C3H6O3\nso the formula is C3H6O3
915044
7.04 grams\n\n15 grams of water is .44 mol and according to the equation, the ratio between water and oxygen is2:1 so if .44 mol of water is decomposed. then .22 mol of oxygen will be produced. and .22 mol of oxygen is .22 *32 =7.04grams
766197
CO2\nCarbon Dioxide\nHere is the reaction:\nHC2H3O2(aq) + NaHCO3(aq) ----> NaC2H3O2(aq) + H2O(l) + CO2(g)
911009
C6H6O + 7O2 yields 3H2O + 6CO2
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?)
825261
First you must balance the equation:\n4NH3 + 5O2  -> 4NO + 6H2O\nthere are 4N, 12H and 10O on each side\nyou have 4.78 moles of O2 you need to divide it by the number of moles of O2 that react and multiply it by the number of moles of NO produced.\n4.78 / 5 x 4 = 3.824moles\nHope this helps
316922
Ok.... \nfirst: 1atm = 760 torr\n Now, as per the gaseous  STP conditions., 1 mole of oxygen will occupy 22.4 lit volume at STP conditions.\nSTP: T=273K\n        P= 760 torr\n       V =22.4 lit\n       Mass of O2 at STP = 32g\n\nNow. second set of conditons:\n P2= 1400 torr\nT2 =41 K\nMass= ?\n\nPV = nRT\nPV = (w/Mwt) XRT\n\nP1/w1T1 = (P2 /W2 T2)\n\nSubstitute these values;;\nW2= 392g of oxygen.\n\nIs the solution is clear enough?
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.
975679
There is no equation.  You have to look at the reaction.\n\nFor example, assume that you have a reaction\n\nC2H4 + 2O2 ---> whatever.\nassuming that the reaction gives you a 100% yield, you need 1 mol of c2H4 and 2 moles of O2.\n\nIf you have 1.5 mol of C2H4, then you will need 3 moles of O2.  If you have only 2.5 moles of O2, then O2 will be the limiting reagents.\n\nYou will use 2.5 moles of O2 and you will need for that 1.25 moles of C2H4.  At the end of the reaction you will have been left with 0.25 moles of C2H4.\n\nWork the other way around if you have 4 moles of O2.  In this case the limiting reactant will be C2H4, because with 1.5 moles of C2H4 you will need only 3 moles O2.  At the end you will be left with 1 mol O2.
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.