KEY Unit 6 - Reaction Stoichiometry and Three Types of Reaction in Aqueous Solution

Transcription

1 KEY Unit 6 - Reaction Stoichiometry and Three Types of Reaction in Aqueous Solution Limitin Reactant and Theoretical Yield Limitin Reaent (or reactant) - the reactant that is completely consumed durin a chemical reaction and limits the amount of product which is formed. Theoretical Yield - the maximum amount of product which can be formed if all of the limitin reaent is consumed and the reaction proceeds to completion. For example: Based on the followin reaction, how many rams of the product (POCl (s) ) can be formed startin with 10.0 rams of each of the reactants? 6 PCl (l) + 6 Cl 2 () + P 4 O 10 (s) > 10 POCl (s) ??? Convert all initial quantities into units usin the correct conversion factors (in this case molar masses). PCl Cl P4 O Determine the limitin reactant usin sound chemical reasonin and calculations. For example, in this case we miht state: The chlorine as cannot be limitin because it is consumed in the same molar quantity as is the phosphorus trichloride but is present in about double the quantity. The choice then is between the phosphorus trichloride and the tetraphosphorus decoxide. Since /6 is less than it is clear that we will run out of the PCl lon before the P 4 O 10 is entirely consumed.... So PCl is the limitin reactant (LR). Find the theoretical yield of product formed by the LR. 15.POCl 10POCl POCl x x pcl 18. 1POCl 6PCl 61 POCl 18.6 of POCl (s) 11

2 Experimental Yield: Percent Yield Even if all reactants in a reaction are at stoichiometry, that reaction often does not o to completion. This can result from: unfavorable conditions reaction bein stopped before it was completed side reactions back reaction leadin to equilibrium Because of the above considerations, the concept of reaction yield is quite important. Actual Yield - the experimentally measured yield of product of a reaction. Percent Yield - the ratio of actual yield to theoretical yield expressed as a percentae. actual yield % Yield x100 theoretical yield Ex: (a) What if the reaction actually yields 18.5 POCl (s)? (b) What if the reaction yields only 1.4 POCl (s)? (a) 99.5 % [see equation above in plu in numbers!!!!] (b) 72.0 % YOU CAN ALSO TRY THESE: (1) 10.0 of aluminum (Al) reacts with 50.0 of sulfur (S) to produce aluminum sulfide (I'll let you determine the formula of this compound and write the balanced chemical equation). The reaction may be written in two forms (both will yield the same answers provided the proper conversion factors are used: 2 Al (s) + S (s) Al 2 S (s) or 16 Al (s) + S 8 (s) 8 Al 2 S (s) (a) What is the limitin reactant? for Al we find Al for S we find 50.0 ether S or S 8 therefore the limitin reactant is Al (s) b/c 1.5*( )<

3 (b) What is the theoretical yield of aluminum sulfide? We et one of aluminum sulfide for every two s of aluminum metal So we find that we can produce a maximum of Al 2 S (s)... and this corresponds to Al 2 S (s) (c) How many rams of the nonlimitin reactant would react and how much would remain unreacted? If of Al 2 S (s) contains 10.0 of Al, then it must contain S. Which means that ( ) of S must remain. This calculation relies on mass conservation. We could also find the number of s of S consumed (1.5 times the number of s of Al consumed) and then convertin that to mass units of sulfur. ( S / 2 Al )*( Al) S S (d) Verify mass conservation. see (c) above and find that: 50.0 S Al yields Al 2 S S (or S 8 ) left-overs (e) What is the % yield if the actual yield is 25.0 of aluminum sulfide? % yield 100*( 25.0 / ) % (a) Al (b) 27.8 (c) 17.8 S react and 2.2 does not react (d) (e) 89.9 % (2) Given the followin reaction, find the theoretical yield of both products. Also, calculate the % yield if 5.92 of Pt(NH ) 2 Cl 2 is actually formed. Based on this % yield, how much KCl would you expect was actually formed? K 2 PtCl 4 (s) + 2 NH () > 2 KCl (s) + Pt(NH ) 2 Cl ?????? follow the same type of loic employed above to find that.06 of KCl and 6.15 of Pt(NH ) 2 Cl % yield 2.95 KCl Quantitative Solution Concentrations You can try these: 1. (a) How many rams of NaCl are needed to prepare 5.00 x 10 2 ml of M NaCl? (b) What volume of this solution will contain s of NaCl? (c) What volume of M NaCl is needed to prepare 75.0 ml of M NaCl? 115

4 (a) 7.1 of NaCl ( NaCl / 1 NaCl )( / Lsol n )(0.500L) (b) 2.00 x 10 2 ml of M NaCl ( 1000 ml sol n / NaCl )*(0.050 ) (c) 45.0 ml of M NaCl (75.0 ml*0.150m) / (0.250 M) Solution Stoichiometry We can now determine the number of s of a substance from information about its concentration in a solution and the volume of the solution. This ives us a new way to et information about limitin reactants in reaction stoichiometry problems. For example, determine the maximum mass of iron (III) carbonate precipitate that can be produced from the quantities of reactants shown below. Na 2 CO (aq) + 2 Fe(NO ) (aq) > Fe 2 (CO ) (s) + 6 NaNO (aq) ml M ml M??? Helpful info.: The molar mass of Fe 2 (CO ) (s) / 9.72 Fe 2 (CO ) (s) (200.0 ml)(0.500 M) 100 ms sodium carbonate (200.0 ml)(0.500 M) 100 ms iron (III) nitrate since we require only 2 s the iron reaent for every s of the carbonate source, the carbonate will run out before all of the iron is precipitated mass ppt ( iron (III) carbonate / 1 ) ( 1 ppt / carbonate)*0.100 mass ppt 9.72 As a follow-up question we could very well ask... How many ml of M Fe(NO ) (aq) are required to completely react with ml of M Na 2 CO (aq)? 1 ml Fe(NO ) (aq)... ( 1000 ml iron sol n / Fe+ )*( 2 Fe / carbonate )* carbonate 116

5 Ex.: How many ml of M NaOH (aq) is required to completely neutralize: (a) 40.0 ml of M H 2 SO 4 (aq) (b) 40.0 ml of M H PO 4 (aq) (a) 20.0 ml of M NaOH (aq) write balanced chemical equation showin that you need 2 NaOH for each H 2 SO 4 2 NaOH (aq) + H 2 SO 4 (aq) 2 H 2 O (l) + Na 2 SO 4 (aq) then do the math to find the ms of acid to be neutralized ( m H2SO4 / 1mL )*40.0mL 2.00 m of acid and the s and ml of base needed to et the job done m of base ( 2 m NaOH / 1 m H2SO4 )*2.00 m acid 4.00 m ml of base ( 1mL NaOH / )*4.00 m NaOH 20.0 ml of the base (b) 0.0 ml of M NaOH (aq) write balanced chemical equation showin that you need NaOH for each H PO 4 NaOH (aq) + H PO 4 (aq) H 2 O (l) + Na PO 4 (aq) then do the math to find the ms of acid to be neutralized ( m HPO4 / 1mL )*40.0mL 2.00 m of acid and the s and ml of base needed to et the job done m of base ( m NaOH / 1 m HPO4 )*2.00 m acid 6.00 m ml of base ( 1mL NaOH / )*6.00 m NaOH 0.0 ml of the base Ex.: 5.67 ml of M HCl (aq) is required to neutralize ml of a NaOH (aq) solution of unknown concentration. What is [NaOH]? Ans: [NaOH] M I ll leave this one to the students... after all, you ve done this in the lab this semester. 117