Homework Help is Here – Start Your Trial Now!
Science
Chemistry
Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion. If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 3.577 Gg silver nitrate reacts with EXCESS of aluminum 2. N: (g) + 3 H₂(g) → 2 NHs (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 N₂ LR 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1g H₂ 1 x 10² g H₂ x x 1 mol N₂ x 2 mol NH3 x 17.04 g NH₂ = 0.75168 g NH; ******* THEORETICAL YIELD 28.02 g N₂ 1 mol N₂ 1 mol NH3 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol NH3 How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N₂ x 1g N₂ 1 x 10² cg N₂ Amount of H₂ Remaining in the Container-H: amount given x 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 0.13366 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ H: amount USED 0.61802 g H₂ GIVEN 0.13366 g H₂ USED = 0.48436 g of H:--LEFT OVER = EXCESS

Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion. If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 3.577 Gg silver nitrate reacts with EXCESS of aluminum 2. N: (g) + 3 H₂(g) → 2 NHs (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 N₂ LR 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1g H₂ 1 x 10² g H₂ x x 1 mol N₂ x 2 mol NH3 x 17.04 g NH₂ = 0.75168 g NH; ******* THEORETICAL YIELD 28.02 g N₂ 1 mol N₂ 1 mol NH3 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol NH3 How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N₂ x 1g N₂ 1 x 10² cg N₂ Amount of H₂ Remaining in the Container-H: amount given x 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 0.13366 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ H: amount USED 0.61802 g H₂ GIVEN 0.13366 g H₂ USED = 0.48436 g of H:--LEFT OVER = EXCESS

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter6: Chemical Reactions: An Introduction
Section: Chapter Questions
Problem 68AP
See similar textbooks
icon
Related questions
Question

FOLLOW THE FORMAT USED IN THE EXAMPLE!

Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion. If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 3.577 Gg silver nitrate reacts with EXCESS of aluminum 2. N: (g) + 3 H₂(g) → 2 NHs (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 N₂ LR 1 x 10² cg N₂ x ? g NH3 = 61.802 cg H₂ x 1g H₂ x 1 x 10² g H₂ 1 mol N₂ x 28.02 g N₂ 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH₂ = 0.75168 g NH; ******* THEORETICAL YIELD 1 mol N₂ 1 mol NH3 2 mol NH3 x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol NH3 How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N₂ x 1g N₂ 1 x 10² cg N₂ Amount of H₂ Remaining in the Container-H: amount given H₂ amount USED . x 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 0.13366 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED 0.48436 g of H₂--LEFT OVER = EXCESS
Transcribed Image Text:Write the balanced equation for the following situation. List the reaction type. Tell the amounts of every substance that remains in the container at the end of the reaction. Assume that all reactions go to completion. If only stoichiometry, tell how much of the excess reactant is used!!!! Reaction Type: a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction Follow the format used in the image. 3.577 Gg silver nitrate reacts with EXCESS of aluminum 2. N: (g) + 3 H₂(g) → 2 NHs (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 N₂ LR 1 x 10² cg N₂ x ? g NH3 = 61.802 cg H₂ x 1g H₂ x 1 x 10² g H₂ 1 mol N₂ x 28.02 g N₂ 1 mol H₂ x 2.02 g H₂ 2 mol NH3 x 17.04 g NH₂ = 0.75168 g NH; ******* THEORETICAL YIELD 1 mol N₂ 1 mol NH3 2 mol NH3 x 17.04 g NH3 = 3.3756 g NH3 3 mol H₂ 1 mol NH3 How much N₂ remains in the vessel? You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED= 61.802 cg N₂ x 1g N₂ 1 x 10² cg N₂ Amount of H₂ Remaining in the Container-H: amount given H₂ amount USED . x 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 0.13366 g H₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 0.61802 g H₂ GIVEN 0.13366 g H₂ USED 0.48436 g of H₂--LEFT OVER = EXCESS
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Stoichiometry
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
Recommended textbooks for you
Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
Chemistry
ISBN:
9781337399425
Author:
Steven S. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Living by Chemistry
Living by Chemistry
Chemistry
ISBN:
9781464142314
Author:
Angelica M. Stacy
Publisher:
W. H. Freeman
Chemistry: Matter and Change
Chemistry: Matter and Change
Chemistry
ISBN:
9780078746376
Author:
Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:
Glencoe/McGraw-Hill School Pub Co
Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
Chemistry
ISBN:
9781305079243
Author:
Steven S. Zumdahl, Susan A. Zumdahl
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781133611097
Author:
Steven S. Zumdahl
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS