PRACTICE WORK-49: STOICHIOMETRY Limiting Reagents, Theoretical Yield, and Percent Yield-01. ) 100 = x85.6%x

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1 CHEM 0 (BEAMER) Last Name Do Date First Name Section M T W R PRACTICE WORK-49: STOICHIOMETRY Limiting Reagents, Theoretical Yield, and Percent Yield-0 General Information Do Date: 5/2 You will need the periodic table, Notes g/mol g/mol g/mol Reaction : Ba + Cl2 BaCl2 ) Student A performs Reaction above. He calculates that the theoretical yield is supposed to be 8.0 grams of barium chloride. His experimental yield is 5.4 grams. Calculate the percent yield. %yield = ( expt theor ) 00 = (5.4 g BaCl g BaCl 2 ) 00 = x85.6%x 2A) Student B performs Reaction and obtains a percent yield of 97.2%. Write this percentage as a conversion factor. (See Notes-55) x85.6 g BaCl 2 expt = 00 g BaCl 2 theorx 2B) If the actual yield was 42. grams, use the conversion factor in 2a to determine the theoretical yield. ( 42. g BaCl 2 (actual) ) ( 00 g BaCl 2 theor 85.6 g BaCl 2 actual ) = x49. g BaCl 2 (theor) %x 3A) Student C performs Reaction and obtains a percent yield of 87.%. Write this percentage as a conversion factor. x87. g BaCl 2 expt = 00 g BaCl 2 theorx 3B) If the theoretical yield is 22.3 g BaCl2, use the conversion factor from 3a to determine the actual yield. ( 22.3 g BaCl 2 (theor) ) ( 00 g BaCl 2 expt 87. g BaCl 2 theor ) = x243.7 g BaCl 2 (expt) %x Page of 6

2 CHEM 0 (BEAMER) INFORMATION FOR QUESTIONS 4A 4E molar mass g/mol g/mol g/mol 2 Cl2 + 3 Al 2 AlCl3 4A) How much aluminum chloride (grams) can be obtained from 0.00 grams of chlorine gas and 0.00 grams of aluminum metal? (This is another way of asking for the theoretical yield of aluminum chloride.) Use initial Cl 2 to determine AlCl 3 : ( 0.00 g Cl 2 ) ( mol Cl 2 ) ( 2 mol AlCl 3 ) ( g AlCl 3 ) = x8.8 g AlCl g Cl 2 2 mol Cl 2 mol AlCl 3 x g Al mol Al Use initial Al to determine AlCl 3 : ( ) ( g Al ) (3 mol AlCl 3 ) ( g AlCl 3 ) = x74.3 g AlCl 2 mol Cl 2 mol AlCl 3 x 3 4B) What is the theoretical yield? 8.8 g AlCl 3 4C) Which substance is the limiting reagent? (LR) Cl 2 4D) Which substance is the reactant in excess (XS) Al 4E) A student performs this experiment, and obtains 6 grams of aluminum chloride. Calculate the percent yield. %yield = ( expt theor ) 00 = ( 6 g BaCl g BaCl 2 ) 00 = x85.6%x Page 2 of 6

3 CHEM 0 (BEAMER) INFORMATION FOR QUESTIONS 5A 5D molar mass g/mol 2.02 g/mol g/mol P4 + 6 H2 4 PH3 5A) How much phosphine gas will be formed from 20. grams of phosphorus (P4) and 0. grams of hydrogen? Initial Value known: g mol mol ratio UNK: mol g LR P4: ( 20. g P 4 ) ( mol P 4 ) ( 4 mol PH 3 ) ( g PH 3 ) = 22 g PH g P 4 mol P 4 mol PH 3 XS H2: ( 0. g H 2 ) ( mol H 2 ) ( 4 mol PH 3 ) ( g PH 3 ) = 2.02 g H 2 6 mol H 2 mol PH g PH 3 (0 g PH 3) 5B) What is the theoretical yield? 22 g PH 3 5C) Which substance is the limiting reagent? (LR) P 4 5D) Which substance is the reactant in excess (XS) H 2 5E) A student performs this experiment, and obtains 5.72 grams of aluminum chloride. Calculate the percent yield. %yield = ( expt theor ) 00 = (5.72 g PH 3 22 g PH 3 ) 00 = x7%x Page 3 of 6

4 CHEM 0 (BEAMER) INFORMATION FOR QUESTIONS 6A 6C molar mass g/mol g/mol g/mol 3 S8 + 6 V 8 V2S3 6A) A student performs this experiment and obtains a percent yield of 94.4%. If the experimental yield is 23.8 grams of vanadium(iii) sulfide, calculate the theoretical yield of vanadium(iii) sulfide. (Another way to say this is: Calculate the mass of vanadium(iii) sulfide that the student should have gotten. ) Hint: See Questions through 3 on this packet. Max Time = 6 min. Max Time on Exam = 4 min The conversion factor should be written: 94.4 g V 2S 3 (expt) = 00 g V 2S 3 (theor) expt yield %-yield conversion factor ( 23.8 g V 2S 3 (expt) ) ( 00 g V 2S 3 (theor) ) 94.4 g V 2 S 3 (expt) = g V 2S 3 6B) Assume that we are still using the information from Question 6a. Assume that vanadium is the limiting reagent. Calculate the initial mass of vanadium. Initial Value known: g mol mol ratio UNK: mol g ( g V 2S 3 ) ( mol V 2S 3 6 mol V ) ( ) ( g V g V 2 S 3 8 mol V 2 S 3 mol V ) = 2.63 g V Page 4 of 6

5 CHEM 0 (BEAMER) INFORMATION FOR QUESTIONS 6A 6C (RE-COPIED HERE FOR YOUR CONVENIENCE) molar mass g/mol g/mol g/mol 3 S8 + 6 V 8 V2S3 6C) CHALLENGE PROBLEM: Assuming that we are still using the information from Question 6a. Assume that sulfur was the reactant in XS. Calculate the mass of leftover S8, assuming that the student began with grams of sulfur. Hints: For the first part of this question, use the theoretical yield of V2S3 to determine the initial amount of sulfur used. For the second part of this question, you will need to do simple subtraction. Initial Value of LR known: g mol mol ratio UNK: mol g S 8 that is used up S 8 that is used up ( g V 2S 3 ) ( mol V 2S 3 ) ( 3 mol S 8 ) ( g S 8 ) =.92 g S g V 2 S 3 8 mol V 2 S 3 mol S 8 Leftover S 8: initial mass used up mass = g S 8.92 g S g S 8 leftover Page 5 of 6

6 CHEM 0 (BEAMER) Answers ) 85.6% 2a) 97.2 g BaCl2 expt = 00 g BaCl2 theor 2b) 43.3 g BaCl2 3a) 87. g BaCl2 actual = 00 g BaCl2 theor 3b) 84.9 g BaCl2 4a) 8.8 g AlCl3 4b) 8.8 g AlCl3. (Yes, this is the same answer as 4a. It s just another way of wording it.) 4c) Cl2 4d) Al 4e) 83.57% 5a) 22 g PH3 5b) 22 g PH3 (Yes, this is the same answer as 5a. It s just another way of wording it.) 5c) P4 5d) PH3 5e) 73% 7% 6a) g V2S3 6b) 2.63 g V 6c) g.93 g = 8.02 g S8 leftover g.92 g = 8.08 g S8 leftover A solutions set will be posted. Page 6 of 6