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Why is part D only 0.2mol? Why can't 0.4mol be made?

Why is part D only 0.2mol? Why can't 0.4mol be made?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Why is part D only 0.2mol? Why can't 0.4mol be made? 

The synthesis of phenylmagnesium bromide and its reaction with methyl benzoate gives triphenylmethanol and biphenyl as major products. A student started with 10.8 g Mg, 62.8 g bromobenzene, 21.8 g methyl benzoate and got 30.0 g of triphenylmethanol and 2 g of biphenyl as products. (Molecular or g atomic weights: Mg, 24.1; bromobenzene, 157.0; methyl benzoate, 136.2; biphenyl, 154.1; triphenylmethanol, 260.3.) Be sure to show your work to answer the following questions. Ryn 1: PhBrt mg → Phmg Br Rxn 2: 2 Ph Mg Br + Ph-eo-0elta (a) What was the limiting reagent in the formation of triphenylmethanol? > Phze-o Cafter hydro bpo) Here 0.45 mol Mg, 0.40 mol Ph Br, ko.16 mol melyl benzote used. Therefore, melhyl benzoake is rhe limiting reagnt (b) What was the limiting reagent in the formation of biphenyl? Bromoben zene -Br Ey mg Br (c) What was the theoretical and % yield of triphenylmethanol? Basad melige benzock, The limiting reagent 6-16 mol x 260-3 g/mol 41.65g Theomhical qicld ソ。Yicld 30-09 ×lo0 = 72ン 41.65g (d) What was the theoretical and % yield of biphenyl? Ph-Br t Ph- Mg Br > Ph-Ph ,', 0-40 mol of PL Br Cam form a mayoimum of 0.2 mul biphenyl. 0-2 mol x 154.1g/mol = 3o.82 g biphenyl – Therischical yield % 4ield. = 6.5% ×1o0 ニ 4 30.82g
Transcribed Image Text:The synthesis of phenylmagnesium bromide and its reaction with methyl benzoate gives triphenylmethanol and biphenyl as major products. A student started with 10.8 g Mg, 62.8 g bromobenzene, 21.8 g methyl benzoate and got 30.0 g of triphenylmethanol and 2 g of biphenyl as products. (Molecular or g atomic weights: Mg, 24.1; bromobenzene, 157.0; methyl benzoate, 136.2; biphenyl, 154.1; triphenylmethanol, 260.3.) Be sure to show your work to answer the following questions. Ryn 1: PhBrt mg → Phmg Br Rxn 2: 2 Ph Mg Br + Ph-eo-0elta (a) What was the limiting reagent in the formation of triphenylmethanol? > Phze-o Cafter hydro bpo) Here 0.45 mol Mg, 0.40 mol Ph Br, ko.16 mol melyl benzote used. Therefore, melhyl benzoake is rhe limiting reagnt (b) What was the limiting reagent in the formation of biphenyl? Bromoben zene -Br Ey mg Br (c) What was the theoretical and % yield of triphenylmethanol? Basad melige benzock, The limiting reagent 6-16 mol x 260-3 g/mol 41.65g Theomhical qicld ソ。Yicld 30-09 ×lo0 = 72ン 41.65g (d) What was the theoretical and % yield of biphenyl? Ph-Br t Ph- Mg Br > Ph-Ph ,', 0-40 mol of PL Br Cam form a mayoimum of 0.2 mul biphenyl. 0-2 mol x 154.1g/mol = 3o.82 g biphenyl – Therischical yield % 4ield. = 6.5% ×1o0 ニ 4 30.82g
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