Presentation is loading. Please wait.

Presentation is loading. Please wait.

Limiting Reagent  What happens in a chemical reaction, if there is an insufficient amount of one reactant?

Similar presentations


Presentation on theme: "Limiting Reagent  What happens in a chemical reaction, if there is an insufficient amount of one reactant?"— Presentation transcript:

1 Limiting Reagent  What happens in a chemical reaction, if there is an insufficient amount of one reactant?

2 In the laboratory, a reaction is rarely carried out with exactly the required amounts of each of the reactants. In most cases, one or more reactants is present in excess. Once one of the reactants is used up, no more product can be formed.

3  Limiting Reagent: the reagent that is completely used up in a chemical reaction.  Excess Reagent: reagent not completely used up in a chemical reaction.

4 Example C + O 2 CO 2 According to the balanced equation, 1 mole of carbon reacts with 1 mole of oxygen to form 1 mole of carbon dioxide. Suppose you mix 5 mole of carbon with 10 mole of oxygen and allow the reaction to take place. In this case there is more oxygen than is needed to react with carbon. Carbon is the limiting reagent and limits the amount of carbon dioxide formed.

5 Example: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O.

6 Solution  Use stoichiometry to calculate one of the products.  The reagent that gave the smaller calculated value of product is the limiting reagent.  The actual value of the amount of product formed is the smaller of the calculated values.

7 Example: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O. Use the grams to mole stoichiometry solution: Calculate H 2 O moles produced by using each of the reactants:  Using O 2 : 0.076 mol H 2 O.  Using H 2 : 0.5 mol H 2 O. The actual amount H 2 O produced is the smaller one of the two values(0.076 mol H 2 O). O 2 is the limiting reagent, since O 2 was used in the calculation of the 0.076 mol H 2 O.

8 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? 4 Na (s) + O 2(g)  2 Na 2 O (s) Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 6.74 g of Na 2 O 5.00g Na ( 1 mole Na ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 6.74 g of Na 2 O 23 g Na 4 mole Na 1 mol Na 2 O 19.38 g of Na 2 O 5.00g O 2 ( 1 mole O 2 ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 19.38 g of Na 2 O 32 g O 2 1 mole O 2 1 mol Na 2 O 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

9 LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4 Na (s) + O 2(g)  2 Na 2 O (s) Use stoichiometry to compare the two reactants. The amount of O 2 used to make 6.74 g of Na 2 O is calculated by: 1.74 g of O 2 was used 5.00g Na ( 1 mole Na ) ( 1 mole O 2 )( 32 g O 2 ) = 1.74 g of O 2 was used 23 g Na 4 mole Na 1 mol O 2 The amount of oxygen (O 2 ) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g in excess 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.

10 LIMITING REACTANT How many grams of solid are formed when 10.0 g of lead reacts with 10.0 g of phosphoric acid? 3 Pb+2 H 3 PO 4  Pb 3 (PO 4 ) 2 (s) + 3 H 2 (g) Start with what is given but do it twice: 13.1 g Pb 3 (PO 4 ) 2 10.0g Pb ( 1 mole Pb ) ( 1 mole Pb 3 (PO 4 ) 2 )( 811 g Pb 3 (PO 4 ) 2 ) = 13.1 g Pb 3 (PO 4 ) 2 207 g 3 mole Pb 1 mole Pb 3 (PO 4 ) 2 41.4 g P b 3 (PO 4 ) 2 10.0g H 3 PO 4 ( 1 mole H 3 PO 4 )( 1 mol Pb 3 (PO 4 ) 2 )( 811 g Pb 3 (PO 4 ) 2 )= 41.4 g P b 3 (PO 4 ) 2 98 g 2 mole H 3 PO 4 1 mole Pb 3 (PO 4 ) 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

11 PRACTICE PROBLEMS 1. How many grams of AgCl product will be produced if 100.00 g of BaCl 2 reacted with excess AgNO 3 ? 2. How many moles of carbon dioxide could be produced from 220.0 g of C 2 H 2 and 545.0 g of O 2 ? 3. How many grams of CO 2 can be produced by the reaction of 35.5 grams of C 2 H 2 and 45.9 grams of O 2 ? 4. In the reaction between CH 4 and O 2, if 25.0 g of CO 2 are produced, what is the minimum amount of each reactant needed? 5. Cu + 2 AgNO 3  Cu(NO 3 ) 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 50.5 g 13.6 mol 9.09 g of CH 4 & 36.4 g of O 2 88.3%

12 PRACTICE PROBLEMS ______1. Which reactant will produce the least amount of AgCl product if reacted with AgNO 3 ? a)100.00 g BaCl 2 b) 400.0 g NaClc) 200.0 g CsCl ______2. How many moles of CO 2 can be produced by the reaction of 5.0 grams of C 2 H 4 and 12.0 grams of O 2 ? ______3. How many grams of carbon dioxide could be produced from 2.0 g of C 2 H 4 and 5.0 g of O 2 ? ______4. In the reaction between CH 4 and O 2, if 18.0 g of CO 2 are produced, how many grams of water are produced? ______5. Cu + 2 AgNO 3  Cu(NO 3 ) 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?


Download ppt "Limiting Reagent  What happens in a chemical reaction, if there is an insufficient amount of one reactant?"

Similar presentations


Ads by Google