Lots of students forget to do the second part (the 15 minus part) and so get graded down. Here's a nice limiting reagent problem we will use for discussion. 0.6543738 mol times 56.1049 g/mol = 36.7 g (to thee sig figs) 4.44 g / 44.009 g/mol = 0.10088845 mol 0.39586 mol − 0.18531 mol = 0.21055 mol Now that we know the limiting reagent is water, this problem becomes "How much H2S is produced from 10.00 g of H2O and excess aluminum sulfide?" Look at the balanced equation for the reaction. 2NaCl(s) + 2NH3(g) + CO2(g) + H2O(ℓ) ---> 2NH4Cl(aq) + Na2CO3(s) Worked example: Calculating the amount of product formed from a limiting reactant Worked example: Relating reaction stoichiometry and the ideal gas law Practice: Stoichiometry: Mental math practice 4) The lowest number indicates the limiting reagent. water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol This means the Al2S3 amount is one-sixth the water value = 0.09251447 mol If the limiting reactant is fully consumed, the reaction will stop even if the other reactant still remains unreacted. : 1. For example, if 1.5 mol C6H6 is present, 11.25 mol O2 is required: 1.5 mol C6H6 x 15 m o l O 2 2 m o l C 6 H 6 = 11.25 mol O2.   I have 30 of them. The value of 0.083 is the important thing. C4H8 + 6O2 ---> 4CO2 + 4H2O (0.02584534 mol) (31.998 g/mol) = 0.827 g of O2 ––– 2) Determine moles of ozone remaining: The test tubes are limiting (they ran out first) and the stoppers are in excess (we have some left over when the limiting reagent ran out). Example #1: Here's a nice limiting reagent problem we will use for discussion. 0.1388 m o l C 6 H 12 O 6 × 6 m o l O 2 1 m o l C 6 H 12 O 6 = 0.8328 m o l O 2. 2) Determine the starting mass of H3PO4 2) Determine how much oxygen reacts with 28 C4H8 molecules: 38 minus 28 = 10 oxygen "groupings" remain after the butane is used up, Example #6: Determine the maximum mass of TiCl4 that can be obtained from 35.0 g of TiO2, 45.0 g Cl2 and 11.0 g of C. (See comment below problem.).   Real Life Examples If you want to put tires on cars and you have 8 cars with no wheels and 48 tires, then cars will be your limiting reagent as you need 4 tires per car. Solution:   see … Solution for limiting reagent, part (a): 6) To solve part (b), we observe that 0.008565 mol of BaO2 was used. 38 minus 28 = 10 oxygen "groupings" remain after the butane is used up x = 0.151332 mol 2C6H10 + 17O2 ---> 12CO2 + 10H2O As the name implies, the limiting reagent limits or determines the amount of product that can be formed. Na2B4O7 is the overall limiting reagent in this problem. 0.091814 / 1 = 0.091814 We will use the amount of water to calculate how much Al2S3 reacts, then subtract that amount from 15.00 g. 2) Use molar ratios to determine moles of Al2S3 that reacts with the above amount of water. Answer: determine the limiting reagent between the first two: 1 Example #7: Determine the starting mass of each reactant if 46.3 of K3PO4 is produced and 92.8 of H3PO4 remains unreacted. Therefore, the limiting reactant in this example is oxygen. of B react, how much of the excess compound remains. The concept of limiting reactants applies to reactions carried out in solution as well as to reactions involving pure substances. Those mole amounts could be used in the calculation below and the final answer could then be multiplied by Avogadro's Number to obtain the answer of 60. The answerer focused on the non-realistic nature of the above chemical equation. The reactant that produces the lesser of the two amounts will tell you the limiting reactant.  = 0.68688 mol TiCl4   Reactant B is a stopper. Cl2 makes the least amount of TiCl4, so Cl2 is the limiting reactant. 79.8658 g TiO2 The reactant that produces the lesser amount of oxygen is the limiting reagent and that lesser amount will be the answer to the question. Stoichiometry will be used to create a ratio between reactants and products given in the balanced chemical equation.   (b) How much of the excess reagent remains unreacted? Since the amount of product produced by oxygen is less than that produced by ammonia, oxygen is the limiting reactant and ammonia is in excess. For the example in the previous paragraph, complete reaction of the hydrogen would yield Be aware!     3) The water is the lesser amount; it is the limiting reagent. How many grams of NO are formed? 0.2181246 mol times 97.9937 g/mol = 21.4 g (to three sig figs) H2SO4 ---> 0.05097 / 1 = 0.05097 (0.02584534 mol) (31.998 g/mol) = 0.827 g of O2, (0.151332 mol) (31.998 g/mol) = 4.84 g of O2. Post was not sent - check your email addresses! You will see the word "excess" used in this section and in the problems. Solution: Example #8: Determine the limiting reagent of this reaction: –––––––––– A balanced equation for the reaction is a basic requirement for identifying the limiting reagent even if amounts of reactants are known. It seems to be a simple concept, but it does cause people problems. ––– The key difference between limiting reactant and excess reactant is that the limiting reactant can limit the amount of final product produced, whereas excess reactant has no effect on the amount of final product.. A reactant is a compound that is consumed during a chemical reaction.A chemical reaction involves reactants – some reactants in excess and some in limited amounts. 1) Determine the moles of Al2S3 and H2O water: 0.555093 mol ÷ 6 mol = 0.0925155 Those mole amounts could be used in the calculation below and the final answer could then be multiplied by Avogadro's Number to obtain the answer of 60. Na2B4O7 is the overall limiting reagent in this problem. There was more than enough of it to react with the other reactant(s). O3 ---> 19.0 g / 47.997 g/mol = 0.39586 mol Reactant A is a test tube. O2 is the limiting reagent. 0.10088845 mol   For the mole calculation: Remember, numbers of molecules are just like moles, so treating the 28 and 228 as moles is perfectly acceptable. This means the H2S amount is one-half the water value = 0.2775465 mol. 0.37062 mol BaO2(s) + 2HCl(aq) ---> H2O2(aq) + BaCl2(aq) ––––––––––– 2) Determine how much oxygen reacts with 28 C4H8 molecules: Answer: determine the limiting reagent between the first two: Na2B4O7 is the limiting reagent when compared to H2SO4. The we solve just as we did in part a just above. In this video we look at solving a sample problem. (0.70635 g) (1 mol HCl / 36.46 g HCl) (1 mol H2O2 / 2 mol HCl) (34.0 g H2O2 / 1 mol H2O2) = 0.332 g H2O2 but the question is "How much remained?" ––––––––––– The technique works, so remember it and use it. (b) How much of the excess reactant is left? BaO2 ---> 1.45 g / 169.3 g/mol = 0.008565 mol Example #6: Determine the maximum mass of TiCl4 that can be obtained from 35.0 g of TiO2, 45.0 g Cl2 and 11.0 g of C. (See comment below problem.) O2: 45.0 g / 31.998 g/mol = 1.406 mol From here figure out the grams of AlI3 and you have your answer. 1.20/2 means there are 0.60 "groupings" of 2 and 2.40/3 means there are 0.80 "groupings" of 3. The final answers will appear with the proper number of significant figures. Three moles of KOH are required to produce one mole of K3PO4 The second factor uses a molar ratio from the chemical equation to convert from moles of the reactant to moles of product. 2) Use molar ratios to determine moles of H2S produced from above amount of water. 1) Determine moles of ozone that reacted: Example #5: Based on the balanced equation: Calculate the number of excess reagent units remaining when 28 C4H8 molecules and 228 O2 molecules react?   Note that the "divide moles by coefficient" was not used to determine the limiting reagent. (0.00224 mol) (36.46 g/mol) = 0.0817 g (to three sig figs) 0.2606 mol times 82.145 g/mol = 21.4 g remaining (to three sig figs) For iodine: 2.40 / 3 = 0.80 water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol Limiting reactant is the reactant that limits the amount of a product that can be formed in a chemical reaction.   5) The second part of the question "theoretical yield" depends on finding out the limiting reagent. the butane:oxygen molar ratio is 1:6 x = 0.01713 mol of HCl used up in the reaction 3) Now, compare the "winner" to the third reagent: ––––––––––– (c) How much excess reagent remains after the reaction is complete? Example #2: 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Solution for part (c): C6H10: 35.0 g / 82.145 g/mol = 0.426 mol Example #8: Determine the limiting reagent of this reaction: 1) Convert everything into moles, by dividing each 5.00 g by their respective molar masses: 2) Note that there are three reactants. Al and AlI3 stand in a one-to-one molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI3. C6H10: 0.426 mol / 2 = 0.213 3) Determine grams of ozone remaining:  =  (b) How much of the excess reactant is left? Na2B4O7 is the limiting reagent when compared to H2SO4 –––––––––––––– 3 mole TiCl4 The reactant that produces the lesser amount of oxygen is the limiting reagent and that lesser amount will be the answer to the question. For the CO if you were to use it up completely you would use up 12.7 mols of CO. You need twice as much H2 as CO since their stoichiometric ratio is 1:2.  x  Once you know that, part (b) becomes "How much H2S can be made from the limiting reagent?" The children in this case would be the limiting reagent because there are less children then there are gloves avaliable. Al is the limiting reagent then second, "What is 15.00 minus the amount in the first part?" 1) Convert everything into moles, by dividing each 5.00 g by their respective molar masses: By the way, did you notice that I bolded the technique to find the limiting reagent? oxygen: 228 / 6 = 38 Once we do that, it becomes a stoichiometric calculation. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant. 3) Finally, we have to do a calculation and it will involve the iodine, NOT the aluminum. Why? Be aware! I certainly hope it is something you pay attention to and remember. b. The calculation gives you the answer to "How much reacted?" Al2S3 + 6H2O ---> 2Al(OH)3 + 3H2S The value of 0.083 is the important thing. Obviously (I hope), the other compound is seen to be in excess. I will do a solution assuming KO2 is the limiting reagent, then I will do a solution assuming CO2 is the limiting reagent. 3) Determine how many moles of the excess reagent is used up when the limiting reagent is fully consumed: 0.2606 mol times 82.145 g/mol = 21.4 g remaining (to three sig figs). Seems obvious, doesn't it? Example 2 Two moles of Mg and five moles of O 2 are placed in a reaction vessel, and then the Mg is ignited according to the reaction M g + O 2 → M g O. H2SO4 ---> 0.05097 mol (1.10 atm) (2.55 L) = (n) (0.08206 L atm / mol K) (292 K)  =  2) Determine the limiting reagent: Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant . Answer: One of the simplest ways to identify a limiting reactant is to compare how much product each reactant will produce. BaO2 (the 0.008565) is the lesser amount, so it is the limiting reagent. H2O ---> 0.2775 mol Comment: this question was asked and answered on Yahoo Answers (nope, no link) and the one answer given (besides mine) totally missed the point of the question. (b) water is associated with the two. Al to O3 molar ratio is 2 to 1 Now, in the example problem, we were more or less told which reactant was the limiting … Na2B4O7 ---> 0.02485 / 1 = 0.02485   1 mole TiO2 2) Note that there are three reactants. 1) Since we have grams, we must first convert to moles. 7) Next, we subtract the amount used up from the total amount that was present: –––– There is no need to convert to grams because all three calculations yield moles of the same compound (the TiCl4). Let's try a simple non-chemical example. x 46.3 g / 212.264 g/mol = 0.2181246 mol of K3PO4 2) Let's say that again: 1) The fact that some phosphoric acid remains tells us it is the excess reagent. The 38 above means that there are 38 "groupings" of six oxygen molecules. (1.10 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (292 K) d. To determine "expected yield" of product, multiply the reaction equivalents for the limiting reagent by the stoichiometric factor of the product. 0.18531 mol times 101.961 g/mol = 18.8944 g 17 4) Determine moles of product formed: 6) To solve part (b), we observe that 0.008565 mol of BaO2 was used. 5) Determine grams of product: x = 0.1654 mol of C6H10 consumed The substance that has the smallest answer is the limiting reagent. 45.0 g Cl2 x  Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The we solve just as we did in part a just above. The first technique is discussed as part of the solution to the first example. 1 Here is the balanced equation for the reaction: (a) Which is the limiting reagent? It is used several different ways: (b) 20 grams of A and 20 grams of B react. –––––––––– ––––––––––– So when the test tubes are used up, we have 10 stoppers sitting there unused. Note: the first factor in each case converts grams of each reactant to moles. Solution: Since there is less of the "grouping of 2," it will run out first. 2 How To Find the Limiting Reactant – Limiting Reactant Example, Free Printable Periodic Tables (PDF and PNG), How Fast Would You Have to Go To Make A Red Light Look Green? If the amount of B actually present exceeds the amount required, then B is in excess and A is the limiting reagent. Solution: For the mole calculation: The lower number is iodine, so we have identified the limiting reagent. You have 1 loaf of sliced white bread, and a package of American cheese individually wrapped slices. This chemistry video tutorial provides a basic introduction of limiting reactants. For example, suppose we have 4 bolts and 8 nuts. 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