Homework Help is Here – Start Your Trial Now!
Science
Chemistry
What is the amount of excess reactant left over at the end of the reaction. Use the first image as an example.

What is the amount of excess reactant left over at the end of the reaction. Use the first image as an example.

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...
See similar textbooks
icon
Related questions
Question

What is the amount of excess reactant left over at the end of the reaction. Use the first image as an example.

3. Ca(OH)2 (aq) + H₂SO4 (aq) → + 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H₂SO4 is the Limiting Reactant) Reaction Type: Neutralization Grams of Product (List the Product and the Amount): 0.180 g H₂O and 0.681 g CaSO4 ? g H₂O = 1.839 g Ca(OH)2 x ? g H₂O = 25.0 mL H₂SO4 X 1 mol Ca(OH)2 X 74.10 g Ca(OH)2 2 mol H₂O x 18.02 g H₂O 1 mol Ca(OH)2 1 mol H₂O = 0.8944 g H₂O 1 L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂O 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol H₂O ONLY USE THE Limiting Reactant TO DETERMINE THE AMOUNT OF THE REMAINING PRODUCT(S). ? g H₂O = 25.0 mL H₂SO4 x 1 L H₂SO4 x 0.200 mol H₂SO4 x 1 mol CaSO4 x 136.14 g CaSO4 = 0.681 g CaSO4 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol CaSO4 YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!!
Transcribed Image Text:3. Ca(OH)2 (aq) + H₂SO4 (aq) → + 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H₂SO4 is the Limiting Reactant) Reaction Type: Neutralization Grams of Product (List the Product and the Amount): 0.180 g H₂O and 0.681 g CaSO4 ? g H₂O = 1.839 g Ca(OH)2 x ? g H₂O = 25.0 mL H₂SO4 X 1 mol Ca(OH)2 X 74.10 g Ca(OH)2 2 mol H₂O x 18.02 g H₂O 1 mol Ca(OH)2 1 mol H₂O = 0.8944 g H₂O 1 L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂O 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol H₂O ONLY USE THE Limiting Reactant TO DETERMINE THE AMOUNT OF THE REMAINING PRODUCT(S). ? g H₂O = 25.0 mL H₂SO4 x 1 L H₂SO4 x 0.200 mol H₂SO4 x 1 mol CaSO4 x 136.14 g CaSO4 = 0.681 g CaSO4 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol CaSO4 YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!!
2. N₂ (g) + 3 H₂(g) → 2 NH3 (g) (This is LIMITING REACTANT: N₂ is the Limiting Reactant) Grams of Product (List the Product and the Amount): 0.75168 g NH3 Reaction Type: Combination or Single Displacement (which is also called Oxidation-Reduction) ? g NH3 = 61.802 cg N₂ x 1g №₂ 2 mol NH3 x 17.04 g NH3 1 mol N₂ 1 mol NH3 LR ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1 x 10² cg N₂ ? g H₂ USED = 61.802 cg N₂ x 1 g H₂ 1 x 10² g H₂ 1g N₂ 1 x 10² cg N₂ X X X 1 mol N₂ x 28.02 g N₂ You will use the LIMITING REACTANT and determine how much H2 was USED in the RXN. 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 28.02 g N₂ 1 mol N₂ 1 mol H₂ 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 Amount of H₂ Remaining in the Container = H₂ amount given H₂ amount USED = = 0.75168 g NH3 ******* THEORETICAL YIELD 3.3756 g NH3 0.13366 g H₂ 0.61802 g H₂ GIVEN - 0.13366 g H₂ USED 0.48436 g of H2--LEFT OVER = EXCESS
Transcribed Image Text:2. N₂ (g) + 3 H₂(g) → 2 NH3 (g) (This is LIMITING REACTANT: N₂ is the Limiting Reactant) Grams of Product (List the Product and the Amount): 0.75168 g NH3 Reaction Type: Combination or Single Displacement (which is also called Oxidation-Reduction) ? g NH3 = 61.802 cg N₂ x 1g №₂ 2 mol NH3 x 17.04 g NH3 1 mol N₂ 1 mol NH3 LR ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1 x 10² cg N₂ ? g H₂ USED = 61.802 cg N₂ x 1 g H₂ 1 x 10² g H₂ 1g N₂ 1 x 10² cg N₂ X X X 1 mol N₂ x 28.02 g N₂ You will use the LIMITING REACTANT and determine how much H2 was USED in the RXN. 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 28.02 g N₂ 1 mol N₂ 1 mol H₂ 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 Amount of H₂ Remaining in the Container = H₂ amount given H₂ amount USED = = 0.75168 g NH3 ******* THEORETICAL YIELD 3.3756 g NH3 0.13366 g H₂ 0.61802 g H₂ GIVEN - 0.13366 g H₂ USED 0.48436 g of H2--LEFT OVER = EXCESS
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Molecules
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS