What is the theoretical yield for this reaction under the given conditions?
Q. 1.93 g H_2 is allowed to react with 9.81 g N_2, producing 1.55 g NH_3. 3H_2(g) + N_2(g) ---> 2NH_3(g) A)What is the theoretical yield for this reaction under the given conditions? B) What is the percent yield for this reaction under the given conditions? Thanks guys please explain how to do if you can, thank you!
Asked by Vincent M - Wed Oct 15 03:51:27 2008 - - 1 Answers - 0 Comments

A. Your requirements are 3 moles H2 to 1 mole N2 to yield 2 moles of NH3 1.93 g H2 = 0.965 moles H2 9.81 g N2 = 0.350 moles N2 1.55 g NH3 = 0.0912 moles NH3 The ratio of H2 to N2 is 2.76, but you need 3, so there is not enough H2; H2 is the limiting reagent and will determine the amount of NH3 produced. Each mole of H2 yields 2/3 moles of NH3, so you should get 0.965*(2/3) = 0.643 moles (10.9 g) of NH3, and that is the theoretical yield. The actual yield is 0.0912 moles, so the percent yield is 0.0912/0.643 * 100 = 14.2%
Answered by gp4rts - Wed Oct 15 04:18:22 2008

What is the theoretical yield of a reaction?
Q. What is the theoretical yield of a reaction if 25.0 grams of product were actually produced from a reaction that has a 88% yield?
Asked by basketbabe51 - Mon Nov 5 21:29:47 2007 - - 2 Answers - 0 Comments

A. 25/88% = x/100%
Answered by kentchemistry.com - Mon Nov 5 21:32:22 2007

What could cause the experimental yield to be greater than the theoretical yield?
Q. In the experiment, aspirin was made, but the actual experimental yield was greater than the theoretical yield. What could have gone wrong?
Asked by Rasheena E - Fri Oct 9 16:25:18 2009 - - 1 Answers - 0 Comments

A. Well assuming that you actually made aspirin there are two reasons that your experiemental yield was more than your theoretical yield: 1. Experimental Error. Did you weigh out a little extra of your starting materials? Was your glassware clean? 2. Impurities. Aspirin made in chem lab is not consumer ready because there are impurities. Perhaps you precipitated out some other materials which were by product of the reaction.
Answered by shea - Fri Oct 9 16:34:58 2009

How do you find a theoretical yield in chemistry?
Q. Ok, so I have 25.0 g of nickel carbonyl and then it decomposes into nickel and carbon monoxide. So I need to know the theoretical yield for the amount of nickel I should get... Can anyone show me how to do this because I totally just forgot. --..Test Tomorrow..-
Asked by pirate - Tue Apr 10 18:36:48 2007 - - 2 Answers - 0 Comments

A. Wow, I can help you with this one! Since you said you "forgot" you probably learned it looong before I did because I just did that like last month. Theoretical yield is the amount of a product that you would supposedly get if you did an experiment. You find it using stoichiometry. So... You have 25.0 grams of nickel carbonyl, which decomposes into nickel and carbon monoxide. How much nickel will you get? The first thing you need is to write a balanced equation for that reaction...so, Ni(CO)4 (I had to google that) yields Ni + CO...and the balanced version of that would be: Ni(CO)4 --> Ni + 4CO All you do is convert grams of nickel carbonyl into grams of nickel using mass-mass stoichiometry. 25.0g Ni(CO)4 x (1 mole/molar mass Ni(CO)4) x ( [cont.]
Answered by purplegrl28 - Wed Apr 11 19:47:26 2007

Does the theoretical yield have to less than the mass of the reactan?
Q. I'm calculating the theoretical yeild of a product in a reaction that I believe is 1:1 stoichiometrically. The mass of the reactant is 0.5g and when I calculate the theoretical yield of the product based on this mass, I end up with 0.627g..Is it possible to get a product mass greater than the reactant mass?
Asked by sssarah - Wed Jun 3 20:30:34 2009 - - 1 Answers - 1 Comments

A. You say "reactant" (singular). If it's really the only reactant, the answer is no. If there is another reactant (such as O2) its mass must be included in the reactant mass.
Answered by kirchwey - Wed Jun 3 20:45:07 2009

How do you calculate the theoretical yield of a fischer esterification reaction?
Q. I have one mole of isoamyl alcohol and one mole of acetic acid. Does this mean that my theoretical yield will be 100%. Please help me! Thanks. :-)
Asked by Rasheena - Fri Sep 18 14:02:03 2009 - - 1 Answers - 1 Comments

A. no... theoretical yield is a # based on the amounts of reagents and the balanced equation. % yield = some % = actual yield / theoretical yield x 100% in your case... C5H11-OH + CH3CO2H ---> CH3CO2-C5H11 + H2O the mole ratio is 1:1 for reactants and you have stoichiometric amounts (1:1 mole ratios) since 1 mole alcohol ---> 1 mole ester... your theoretical yield = 1 mole ester.. IF you recovered 1 mole ester in your experiment, your % yield = recovered / theoretical = 1/1 x 100% = 100% IF you recovered 1/2 mole, your % yield = 0.5 / 1 x 100% = 50 % see how that works? questions?
Answered by m w - Fri Sep 18 14:28:24 2009

Can anyone explain how you get theoretical yield in detail please?
Q. I've searched everywhere and nobody really explains the ratios and stuff that are involved in calculating theoretical yield. Please help.
Asked by Jenn M - Thu Sep 3 20:47:57 2009 - - 2 Answers - 0 Comments

A. Theoretical yield is determined using the balanced equation for the reaction. Here is a simple example. 2H2 + O2 === 2H2O If you use only 1/2 mole of O2 then the theoretical yield of H2O is 1 mole. You can also use masses. If 2 g of H2 is used (1 mole) then 18 g of H2O (1 mole) will be produced. These are theoretical yields. In an experiment you may actually obtain less.
Answered by Rick - Thu Sep 3 21:01:47 2009

What is the theoretical yield of calcium carbonate?
Q. If 2.16 grams of calcium chloride dihydrate was dissolved in water and added to sodium carbonate (in excess), what is the theoretical yield (grams) of calcium carbonate?
Asked by Tony - Sun Mar 29 14:21:09 2009 - - 1 Answers - 0 Comments

How to calculate the theoretical yield of this problem?
Q. This is the problem: The decomposition reaction of calcium carbonate is represented by the following balanced equation: CaCO3(s) => CaO(s) + CO2(g) A 15.8-g sample of calcium carbonate (CaCO3) was heated in an open container to cause decomposition. Calculate the theoretical yield of CO2 expected to be produced according to the following equation: can someone please help me going step by step with this problem? thanx!
Asked by hello - Thu Jun 25 17:38:55 2009 - - 4 Answers - 0 Comments

A. 15.8 g of CaCO3 (m.w. = 40+12+3*16=100 g) is 0.158 moles so you should leave behind 0.158 moles of CaO (m.w.=40+16 = 56 g) = 8.85 g and generate 0.158 moles of CO2 (m.w. = 12+2*16 = 44 g) = 6.95 g but without the equation we can't figure out more than that.
Answered by unknown - Thu Jun 25 17:52:51 2009

What is the theoretical yield of Fe2O3 in grams ?
Q. Iron reacts with oxygen to form iron (III) oxide according to the reaction : 4Fe(s)+3O2(g) --> 2Fe2O3(s) If you begin with 3.46 of Fe: what is the theoretical yield of Fe2O3 in grams?
Asked by Tyfani - Thu Nov 12 10:37:55 2009 - - 1 Answers - 0 Comments

A. using molar masses, 4Fe(s)+3O2(g) --> 2Fe2O3(s) becomes: 4(55.84g) Fe --> 2(159.69 g)Fe2O3 223.4 g Fe --> 319.4g Fe2O3 If you begin with 3.46 of Fe: what is the theoretical yield of Fe2O3 in grams? 3.46 of Fe @ 319.4 g Fe2O3 / 223.4 g Fe = 4.95 grams of Fe2O3
Answered by Steve O - Thu Nov 12 13:41:40 2009

How do I determine the theoretical yield?
Q. Determine the theoretical yield of C when each of the following amounts of A and B are allowed to react in the generic reaction: 2A + 3B -> 2C Besides the answers I'm requesting you show how you got them which is why I'm posting 2 of the 4 questions I have to solve. Thanks in advance.
Asked by Donna - Fri Oct 23 00:49:41 2009 - - 1 Answers - 0 Comments

When calculating the theoretical yield, do we consider the mole of the solvent used?
Q. When I was calculating the theoretical yield of a product, the solvent turned out to be the limiting reagent meaning thats the smallest mole.. So should I take that as the limiting reagent when I calculate the product yeild? or should I just avoid it and use the smallest mole of one of the reactants?
Asked by Curious Sarah - Sun Apr 26 15:54:23 2009 - - 1 Answers - 0 Comments

A. The solvent is NOT involved in calculating the theoretical yield unless it is one of the reactants. Only the reactants are directly linked to the products. The solvent is simply the medium to allow the transformation take place.
Answered by unknown - Sun Apr 26 16:03:40 2009

What is the theoretical yield of CO2?
Q. The burning of 18 grams of carbon produces 55 grams of carbon dioxide. What is the theoretical yield of CO2? Calculate the percent yield of CO2, I need help with this problem. How do you do it?
Asked by Ashley - Fri Dec 19 02:34:56 2008 - - 2 Answers - 0 Comments

A. thermochemical equation is C+O2---C02 12g carbon GIVES 44g C02 HENCE,18---=18*44/12 =66g Hence theoritical yield=66g and % yield=experimental yield*100/theoritical yield therefore,%yield=55*100/6 6=83% Ihope you can understand it.
Answered by Ankur M - Fri Dec 19 03:11:36 2008

How do I calculate this theoretical yield problem?
Q. Consider the reaction: 2 A + 3 B ---> C + D Molecular Weights: A = 138 g/mol B = 140 g/mol C = 175 g/mol D = 85 g/mol You have 5.2 grams of A and 4.6 grams of B. What is the theoretical yield of C? I used the molar mass of the limiting reagent B times the ratio of C to B then multiplied by Cs molar mass. What am I doing wrong?
Asked by Stacy - Fri Sep 18 17:41:15 2009 - - 2 Answers - 0 Comments

A. Well first you'll need to find the limiting reagent: .0188 moles A (if divided by stoichiometric ratio so essentially moles A / 2) .010952 moles B (Again moles B / 3) Thus B is the limiting reagent. So the theoretical yield would be: 4.6 g B * 1 mol B / 140g B * 1 mol C / 3 mol B * 175 g C / 1 mol C So your solution is 1.92 g of C However there seems to be an issue of conservation of mass in this problem. If it requires 2 moles of A and 3 moles of B to make 1 mole of C and D, the molecular weight of C + D should be far larger (unless i guess mass is converted into energy).
Answered by Kevin H - Sun Sep 20 12:04:03 2009

What is the theoretical yield of CH3OH ?
Q. Methanol, CH3OH, can be made by combining gaseous carbon monoxide and hydrogen gas. In one experiment 1.80 g of H2 is mixed with 15.60 g of CO. What is the theoretical yield of CH3OH ? a.k.a. How many grams of CH3OH should be formed?
Asked by tibsy2007 - Thu Nov 15 01:43:20 2007 - - 4 Answers - 0 Comments

A. CO + 2H2 --> CH3OH CO = 15.6/(16+12)=0.557 mol H2= 1.8/2 = 0.9 mol H2 as limitting reactan CH3OH formed = 1/2*0.9 = 0.45 mol mass = 0.45*(12+4+16) =14.4 g
Answered by rendra - Thu Nov 15 01:53:12 2007

Can someone help me with this problem in the theoretical yield?
Q. A mass of 13.02 g of (NH4)2SO4 is dissolved in water. After the solution is heated, 27.22 g of Al2(SO4)3 * 18H2O is added. Calculate the theoretical yield of the resulting alum. This is a limiting reactant problem.
Asked by daniel b - Wed Apr 8 21:58:51 2009 - - 1 Answers - 0 Comments

A. yeild = .0806
Answered by Young Master Ruprecht - Sun Apr 12 16:52:46 2009

Which reactant is the limiting reagent and what is the theoretical yield?
Q. If the coordination complex, K2 (Cu(C2O4))2H2O is synthesized from 1.55g of copper sulfate and 2.5 g of K2C2O4.H2O. Which reactant is the limiting reagent? What is the theoretical yield? CuSO4.5H2O + 2K2C2O4.H2O --> K2(Cu(C2O4)2).2H2O +K2SO4 + 5H2O
Asked by kiwibaby231 - Fri Feb 20 10:13:45 2009 - - 1 Answers - 0 Comments

A. moles CuSO4*5H2O = 1.55 g / 249.68 g/mol= 0.00621 Moles K2C2O4*H2O = 2.5 g / 184.242 g/mol=0.0136 CuSO4*5H2O is the limiting reactant ( 0.0136/2 = 0.00680 moles are needed) we would get 0.00621 moles of the complex mass = 0.00621 mol x 335.81 g/mol=2.1 g
Answered by Dr.A - Fri Feb 20 10:25:19 2009

What is the theoretical yield of calcium oxide if 24.80g of calcium carbonate decomposed?
Q. use this balanced equation to solve the problem CaCO3(s) CaO(s) + CO2(g) What is the theoretical yield of calcium oxide if 24.80g of calcium carbonate decomposed completely? could you please explain the problem a little thanks sooo much as answer I got .36 CaO Mols is this right ???
Asked by Dani B - Wed Mar 25 23:38:44 2009 - - 1 Answers - 0 Comments

A. The theoretical yield is the maximum possible amount of a product that you can form from the reactants given. So in this case the theoretical yield is however much CaO is possible from 21.80 g of CaCO3. CaCO3(s) CaO(s) + CO2(g) You can see from the balanced equation that 1 mole CaCO3 decomposes to give 1 mole of CaO Therefore the theoretical moles of CaO produced is = moles CaCO3 started with (because 1 mole CaCO3 gives 1 mole CaO moles CaCO3 = mass / molar mass molar mass CaCO3 = 100.09 g/mol moles CaCO3 = 24.80 g / 100.09 g/mol moles CaCO3 = 0.24777 moles You have decomposed 0.24777 moles of CaCO3 This many moles of CaCO3 will give 0.24777 moles of CaO Now, convert 0.24777 moles CaO to mass mass = molar mass x moles molar mass… [cont.]
Answered by Lexi R - Thu Mar 26 03:08:26 2009

* I will give 10 points! * What is the theoretical yield and percent yield of this?
Q. Calculate the theoretical yield of C2H5Cl when 126 g of C2H6 reacts with 224 g of Cl2, assuming that C2H6 and Cl2 react only to form and C2H5Cl. Then, calculate the percent yield if the reaction produces 183 g of C2H5Cl. I will give 10 points to correct answer! (Please answer both questions!) Thank you.
Asked by `~-Cynic - Fri Sep 19 23:16:39 2008 - - 4 Answers - 0 Comments

A. Since I doubt this is a made up problem, I believe there are actually two products, C2H5Cl and HCl. So you don't need to worry about stochiometric coefficients they react in a 1:1 ratio. 1. Calculate mol mass. 1 mol of C2H6 = 30g and 1 mol of Cl2(which is a diatomic gas, so 2 atoms in a molecule) = 71g 2. Find limiting reactant. 126/30=4.2mol of C2H6; 224/71=3.2mol of Cl2 3. Since they react 1:1 you will have 3.2 mol of C2H5Cl which is 65.5g/mol. 65.5*3.2=209.4g and that is your theoretical yield. Take your experimental yield and divide by theoretical to get percent yield: 183/209.4=.87 or 87%
Answered by Duke O - Fri Sep 19 23:55:41 2008

How to find the theoretical yield?
Q. Ok...The question is to find out the %yield..but in order to do that I need to find out the theoretical yield 1NaHCO3 +1 HCl ->1 NaCl +1CO2 + 1 H2O 1Na2CO3+2HCl -> 2NaCl + 1CO2 +1H2O The first one has a 1:1 mole ratio, while the second one has a 1:2 mole ratio. Can someone tell me how to go about finding the theoretical yield?
Asked by Its a Girl! - Sun Nov 1 09:06:44 2009 - - 1 Answers - 0 Comments

A. how many grams (or moles) of NaHCO3 and Na2CO3 do you have initially? From there, by knowing the mole ratios of your reactions, you would be able to know the theoretical yield
Answered by smiley - Sun Nov 1 09:18:33 2009