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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 LR 2 mol NH3 x 17.04 g NH3 = 1 mol N₂ 1 mol NH3 ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1 g N₂ 1 x 10² cg N₂ 1g H₂ 1 x 10² g H₂ X X 1 mol N₂ x 28.02 g N₂ 1 mol H₂ 2.02 g H₂ X 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 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 X 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 1 x 10² cg N₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 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

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 LR 2 mol NH3 x 17.04 g NH3 = 1 mol N₂ 1 mol NH3 ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1 g N₂ 1 x 10² cg N₂ 1g H₂ 1 x 10² g H₂ X X 1 mol N₂ x 28.02 g N₂ 1 mol H₂ 2.02 g H₂ X 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 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 X 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 1 x 10² cg N₂ 28.02 g N₂ 1 mol N₂ 1 mol H₂ 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

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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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. 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) 1 g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 x 1 mol N₂ 17.04 g NH3 = 1 mol NH3 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1 g H₂ 1 x 10² g H₂ ? g H₂ USED = 61.802 cg N₂ x 1 g N₂ 1 x 10² cg N₂ X X X 1 mol N₂ x 28.02 g N₂ 1 mol H₂ 2.02 g H₂ How much N2 remains in the vessel? 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 X 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 0.13366 g H₂ NH3 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) 1 g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 x 1 mol N₂ 17.04 g NH3 = 1 mol NH3 1 x 10² cg N₂ ? g NH3 = 61.802 cg H₂ x 1 g H₂ 1 x 10² g H₂ ? g H₂ USED = 61.802 cg N₂ x 1 g N₂ 1 x 10² cg N₂ X X X 1 mol N₂ x 28.02 g N₂ 1 mol H₂ 2.02 g H₂ How much N2 remains in the vessel? 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 X 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 0.13366 g H₂ NH3 0.61802 g H₂ GIVEN 0.13366 g H2 USED = 0.48436 g of H₂--LEFT OVER = EXCESS
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