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65.37 g Zn YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!! 1 mol Zn 1 mol H₂

65.37 g Zn YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!! 1 mol Zn 1 mol H₂

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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4. Zn (s) + 2 HCl(aq) → ZnCl₂ (aq) + H₂(g) (This is just stoichiometry.) Reaction Type: Single Displacement Grams of Product (List the Product and the Amount): 7.48 x 10 g ZnCl₂ and 1.11 x 10¹ g H₂ ? g ZnCl₂ = 3.59 x 10² g Zn x 1 mol Zn x 65.37 g Zn 1 mol ZnCl₂ x 1 mol Zn 136.27 g ZnCl₂ = 7.48 x 10 g ZnCl₂ 1 mol ZnCh ? g H₂ = 3.59 x 10 g Zn x 1 mol Zn x 2.02 g H₂ = 1.11 x 10¹ g H₂ 1 mol H₂ 65.37 g Zn YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!! 1 mol H₂ x 1 mol Zn What is the amount of excess reactant left over at the end of this reaction? 3.59 x 10^2 g strip of zinc is dropped into a test tube of containing excess hydrochloric acid
Transcribed Image Text:4. Zn (s) + 2 HCl(aq) → ZnCl₂ (aq) + H₂(g) (This is just stoichiometry.) Reaction Type: Single Displacement Grams of Product (List the Product and the Amount): 7.48 x 10 g ZnCl₂ and 1.11 x 10¹ g H₂ ? g ZnCl₂ = 3.59 x 10² g Zn x 1 mol Zn x 65.37 g Zn 1 mol ZnCl₂ x 1 mol Zn 136.27 g ZnCl₂ = 7.48 x 10 g ZnCl₂ 1 mol ZnCh ? g H₂ = 3.59 x 10 g Zn x 1 mol Zn x 2.02 g H₂ = 1.11 x 10¹ g H₂ 1 mol H₂ 65.37 g Zn YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!! 1 mol H₂ x 1 mol Zn What is the amount of excess reactant left over at the end of this reaction? 3.59 x 10^2 g strip of zinc is dropped into a test tube of containing excess hydrochloric acid
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