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2. Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of nitrogen Grams of hydrogen Grams of ammonia

2. Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of nitrogen Grams of hydrogen Grams of ammonia

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
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2. Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of nitrogen Grams of hydrogen Grams of ammonia
Transcribed Image Text:2. Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of nitrogen Grams of hydrogen Grams of ammonia
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) 1g N₂ ? g NH3 = 61.802 cg N2 x LR 1 mol N₂ x 2 mol NH3 x 17.04 g NH3 = 28.02 g N₂ 1 mol N₂ 1 x 10² cg N₂ 1 mol NH3 ? g NH3 = 61.802 cg H₂ x How much N2 remains in the vessel? 1 g H₂ 1 x 10² g H₂ ? g H₂ USED= 61.802 cg N₂ x X X X 1 mol H₂ 2.02 g H₂ x You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 28.02 g N₂ 1 mol N₂ 1 mol H₂ 0.75168 g NH3 ******* THEORETICAL YIELD 3.3756 g NH3 1 g N₂ 1 x 10² cg N2 Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 0.61802 g H₂ GIVEN 0.13366 g H₂ USED = 0.48436 g of H₂--LEFT OVER = EXCESS 0.13366 g H₂
Transcribed Image Text: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) 1g N₂ ? g NH3 = 61.802 cg N2 x LR 1 mol N₂ x 2 mol NH3 x 17.04 g NH3 = 28.02 g N₂ 1 mol N₂ 1 x 10² cg N₂ 1 mol NH3 ? g NH3 = 61.802 cg H₂ x How much N2 remains in the vessel? 1 g H₂ 1 x 10² g H₂ ? g H₂ USED= 61.802 cg N₂ x X X X 1 mol H₂ 2.02 g H₂ x You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 1 mol N₂ x 3 mol H₂ x 2.02 g H₂ = 28.02 g N₂ 1 mol N₂ 1 mol H₂ 0.75168 g NH3 ******* THEORETICAL YIELD 3.3756 g NH3 1 g N₂ 1 x 10² cg N2 Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 0.61802 g H₂ GIVEN 0.13366 g H₂ USED = 0.48436 g of H₂--LEFT OVER = EXCESS 0.13366 g H₂
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