Limiting Reagent confusion?
477943
you need 2 moles of hydrogen for each mole of oxygen so for four oxygen moles you need eight hydrogens. oxygen is the limiting reactant and you can make four water moles
263093
Not limiting properly
445261
Try it yourself, its not that bad. \nHere is how:\n1) Set up a stoichometry ratio.\n2) Using the mass that they give you, calculate the theoretical yield of each product.\n3) Whichever product has a smaller theoretical yield, analyze the reactants which caused it to be such.\n4) The one that 'limits' the amount of products is the limiting reagent.\n5) Using that value, calcuate the amount of the other reactant that is necessary. Subtract from the orginal value to find the excess reactant.
793971
According to the equation:\n\n2NO + O2 --> 2NO2\n\n2 mol of NO reacts with 1 mol O2, so\n0.886 mol NO reacts with 0.886/2 = 0.443 mol O2\n\nWe have 0.503 mol O2 > 0.443 mol. So O2 is in excess and NO is the limiting reactant. It reacts fully and produces 0.886 mol NO2 (because 2 mol NO give 2 mol NO2)
658363
This is a limiting reagent problem. The first job is to figure out how many moles of each reagent you have and then figure out which one will run out first because that will determine how much product you can make. It is easiest to write out the balanced equation first to see what you are dealing with:\n\n2Na + Cl2 --> 2NaCl\n\nYou have 23g x 1mol/23g the molar mass of sodium = 1 mol of sodium\n\nYou plug the standard pressure, 1 atm, the volume, 0.56 L, the R constant, 0.082, and the standard temperature, 273 K, into the ideal gas law formula PV = nRT and solve for n.\n\nYou get 0.0250 moles chlorine gas.\n\nThen figure out how many moles of chlorine are needed for the amount of sodium you have.\n\nfor 1 mol of Na you would need 1 x 1/2 = 0.5 moles of chlorine. But you only have 0.025 moles so this must be the limited reagent.\n\nNow you must figure out how many moles of NaCl form from 0.025 moles of Cl2.\n\n0.025 x 2/1 = 0.05 moles of NaCl.\n\nNow multiply this by the molecular weight for NaCl, 23 g/mol for Na and 35.5 g/mol for Cl = 58.5 g/mol total\n\n0.05 mol x 58.8 g/mol = 2.93 or about 3 grams of NaCl formed
791950
Is this 80/20 by volume, or mole fraction? In the case of volume, just hook up the tanks and regulate flow rate. For mole fraction, I suppose you could just find the volumetric ratio from the mole ratio, factoring in compressibility and such, and then do the same as with volumetric ratio-- hook up the argon and CO2, make argon inlet flow four times as much as CO2 volumetric inlet flow.\n\nAs for if the ratio is wrong? That depends on the process. Argon is an inert gas, so it won't take part in reactions. However, product yield will go down if CO2 is a limiting reactant in the process, or if the reaction rate depends on CO2. Too much CO2 may result in a waste of CO2 in the process, since conversion may depend on the other reactants. Again, without more knowledge of what the process is, it's hard to say exactly what will happen, but there will definitely be money lost, whether in the form of less products, or wasted CO2.
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
423379
let me see if i can at least help you with part of your problem, solubility in chemistry is simple, basically the equation will always be:\n\n[Solubility equalibrium constant] = concentration of product divided by concentration of reagents (concentration of each reagents multiplying by each other).\n\nthe easiest way to understand this is to simplify things for you so as you know, a chemical equation always go like this:\n\nreagents <==> products\n\nJust know this from the above, the dynamic here is that when you have an excess of either the reagents or the product, the balance of the equation tend to shift in the opposite direction (ie excess concentration of product will cause more reagents to form and vice versa), in a chemical equation such as the one from above, the reaction can actually go either way, products or reagents can form.  What really matters is the equalibrium constant that I talked about in the first equation. This constant represent the concentration of the products and the reagents when they are in equailibrium or when they are balance, meaning that the rate at which products or reagent form stays the same and there are no change in the concentration of either the products or the reagents so the whole thing is static at the point.   Think about this, more concentration of products wouldn't form unless you have extra reagents to spare that can shift the balance.  When something is saturated it means that it can longer dissolve in a solvent (let's use water to simplify the explanation)so it just accumulates at the bottom, water molecules can only separate so many ions from each other (the reagents) and preventing them from forming a solid product since water itself has limited molecules (if you mix enough table salt in a glass of water, it'll get to the point where it'll accumulate at the bottom and won't dissolve).  If remember correctly, in a chemical equation, when there's different amount of reagents needed to form the product, there's a slight modification to the equation I gave you, for example:\n\n2 CaCO3 + 4 ZnO2 <----> 3 ZnCO3 + 3 CaO2\n\nthen  [equilibrium constant] = [Cao2]^3 x [ZnCO3]^3 divided by [CaCO3]^2 x [ZnO2]^2\n\nnote that you should balance the chemical equations first, I am just giving an example.\n\nhere's a summary of how to compare numbers, if you plug given concentrations of reagents and product into the equation and the number you get is larger than the real equilibrium constant, then the concentration of products is higher than it should be and the equation will shift towards forming more reagents (in the case of the solid product, it'll dissolve) and if there's more reagents, more solid product is gonna form.  Hope that helps. Remember the Chemistry textbook is still your best reference.
1088842
What is the experiment?\n\nWhat were ALL the reagents?\n\nAs it stands, your question is unanswerable; it is like me asking you what is on my kitchen table.
812946
Chemical reagents are not used to dilute blood. Depending on the test needed, some have a non clotting agent put in. Others let it clot and take off the serum to test for various reasons. What you are probably referring to is an agent to either stop clotting or preserve the specimen.
609634
This question cant be answered without knowing the molarity of the HCl. I seriously doubt you were using 12M as another poster suggested. \n\nAlso. if all the Mg reacted the moles of Mg reacted would simply be 0.15/24.3 = 0.006. (Did you really use 0.15g? Thats a small amount) \n\nIf you didnt actually preform the rxn and/or dont know if all the Mg reacted you need to use the moles of Mg calculated above, and compare it to the moles of HCl used. M = moles / liters. So plug in the molarity of HCl used, and 0.050 L (converted 50mL) and solve for moles of HCl. Next use the mole ratio to determine the limiting reagent. If you dont know how to do that, your screwed.
270274
ur equation is wrong....\nu will get H2O + C...\ncheck again\nif it's correct, u have to find out which one is the limiting reagent by calculating the number of moles of each reactant used...thus figuring out the number of moles of product and multiplying with the molecular mass of methanol...
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.
16164
The instruction is not correct because the total volume yielded when you mix 100ml of concentrated solution and 900ml water will be somewhat less than 1000ml. The concentration will be slightly higher than 1.0M\n\nThe correct procedure is to take a 1l volumetric flask  and add the concentrated solution to about 800ml water, then fill up to the mark.
264121
The theory can be used to calculate the mass of product that is expected to be produced when more than one reactant is involved in its formation.\nConsider the reaction below:\n\n2Na2S + Na2CO3 + 4SO2 ---> 3Na2S2O3 + CO2 \n\nGiven the mass of the reactants involved, we would use just one of them as a reference to get the mass of the product, but which reactant should we choose? \n\nHere we find the moles of each of the 3 reactants (given the mass)  and then compare the moles ratio of each in turn with the product and hence calculate the moles of the product Na2S2O3. A limiting reactant is one which gives the least moles of the product. So we see that from which reactant we get the least moles of Na2S2O3, and then with the moles, we can finally calculate the mass of product we expect to get when the reaction is carried out.\nWe choose the limiting reactant for finding out the yield because after any one of the reactants is used up,  the reaction won't proceed further, so this is to calculate the maximum amount of product we can get.
712891
In one experiment 32.40 g of NO is mixed with 19.20 g of O2.\n\n\nThe reaction goes: 2NO + O2 -> 2NO2\n\n\nFirst, you have to find out which is the limiting reactant.\n\nno of moles of NO\n= mass / molar mass\n= 32.40 / (14 + 16)\n= 32.40 / 30\n= 1.08\n\nno of moles of O2\n= mass / molar mass\n= 19.20 / (2 x 16)\n= 0.6\n\nGoing by the equation, there must be twice the no of moles of NO than O2. However, the no of moles of O2 supplied is more than half of the no of moles of NO. Thus, O2 is in excess and NO is the limiting reactant.\n\nWorking with the no of moles of NO,\nno of moles of NO2\n= no of moles of NO\n= 1.08\n\nmass of NO2\n= (no of moles) x (molar mass of NO2)\n= 1.08 x (14 + 2 x 16)\n= 1.08 x 46\n= 49.68g\n\nAnd there you have it! :D