1 / 6

LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction.

LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction.

daryl
Download Presentation

LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction. To understand this concept, let’s suppose you were an elf working for Santa Claus and your job was to make candy canes. You take one red stick and one white stick then twist them around to make one candy cane. The ratio of red to white is 1:1. Time is ticking and you find that you have 24 red stick and 17 white sticks left. What is the maximum number of candy canes you can make? The answer is 17 candy canes!The white sticks “limit” the amount of product you could make. In chemistry we do not use sticks but *MOLES* to determine which starting material will limit the maximum amount of product that can be produced in a chemical reaction.

  2. LIMITING REACTANT a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4Na (s)+ O2(g)2 Na2O(s) Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 5.00g Na(1 mole Na)(2 mole Na2O)( 62 g Na2O) = 6.74 g of Na2O 23 g Na 4 mole Na 1 mol Na2O 5.00g O2(1 mole O2)(2 mole Na2O)( 62 g Na2O) = 19.38 g of Na2O 32 g O2 1 mole O2 1 mol Na2O Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide. Wrong answer

  3. LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4Na (s)+ O2(g)2 Na2O(s) There are two methods used to answer this question. The Law of Conservation of mass and Stoichiometry. The amount of O2 used to make 6.74 g of Na2O is calculated by: 5.00g Na(1 mole Na)(1 mole O2)( 32 g O2) = 1.74 g of O2 was used 23 g Na 4 mole Na 1 mol O2 Or use the Law of Conservation of mass: Mass of product (6.74 g) – mass of limiting reactant (5.00 g) = mass of other reactant, in this case oxygen (1.74 g). The amount of oxygen (O2) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

  4. LIMITING REACTANT How many grams of solid are formed when 10.0 g of lead reacts with 10.0 g of phosphoric acid? 3Pb+2 H3PO4 Pb3(PO4)2 (s) + 3 H2 (g) Start with what is given but do it twice: 10.0g Pb(1 mole Pb)(1 mole Pb3(PO4)2)(811 g Pb3(PO4)2) = 13.1 g Pb3(PO4)2 207 g 3 mole Pb 1 mole Pb3(PO4)2 10.0g H3PO4(1 mole H3PO4)(1 mol Pb3(PO4)2)(811 g Pb3(PO4)2)=41.4 gPb3(PO4)2 98 g 2 mole H3PO41 mole Pb3(PO4)2 You can not have two different answers for one question so in this case lead “limits” how much lead(II) phosphate that can be produced. The correct answer is 13.1 g. Wrong answer

  5. PRACTICE PROBLEM #19 1. How much AgCl product will be produced if 100.00 g of BaCl2 reacted with excess AgNO3? 2. How many moles of carbon dioxide could be produced from 220.0 g of C2H2 and 545.0 g of O2? 3. How many grams of CO2 can be produced by the reaction of 35.5 grams of C2H2 and 45.9 grams of O2? 4. In the reaction between CH4 and O2, if 25.0 g of CO2 are produced, what is the minimum amount of each reactant needed? 5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag. When 10.0 g of copper was reacted with 60.0 g of silver nitrate solution, 30.0 g of silver was obtained. What is the percent yield of silver obtained? 137.57 g 13.63 mol 50.5 g 9.09 g of CH4 & 36.4 g of O2 88.3%

  6. GROUP STUDY PROBLEM #19 ______1. Which reactant will produce the least amount of AgCl product if reacted with AgNO3? a)100.00 g BaCl2 b) 400.0 g NaCl c) 200.0 g CsCl ______2. How many moles of CO2 can be produced by the reaction of 5.0 grams of C2H4 and 12.0 grams of O2? ______3. How many grams of carbon dioxide could be produced from 2.0 g of C2H4 and 5.0 g of O2? ______4. In the reaction between CH4 and O2, if 18.0 g of CO2 are produced, how many grams of water are produced? ______5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag. When 50.0 g of copper was reacted with 300.0 g of silver nitrate solution, 149 g of silver was obtained. What is the percent yield of silver obtained?

More Related