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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: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter3: Equation, The Mole, And Chemical Formulas
Section: Chapter Questions
Problem 3.49QE: One of the ways to remove nitrogen monoxide gas, a serious source of air pollution, from smokestack...
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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
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