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YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!!

YOU MUST CALCULATE THE AMOUNT OF EXCESS REACTANT LEFT OVER AT THE END OF THE REACTION!!!!!!!

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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Calculate the amount of excess reactant left over at the end of the reaction, use the 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 1 mol Ca(OH)2 x 18.02 g H₂O 1 mol H₂O = 0.8944 g H₂O 1 L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol H₂O = 0.180 g 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 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol CaSO4 = 0.681 g 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 1 mol Ca(OH)2 x 18.02 g H₂O 1 mol H₂O = 0.8944 g H₂O 1 L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol H₂O = 0.180 g 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 1000 mL H₂SO4 1 L H2SO4 1 mol H₂SO4 1 mol CaSO4 = 0.681 g 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) 1 g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 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 How much N₂ remains in the vessel? 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 You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. 1 mol H₂ x 2.02 g H₂ 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 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) 1 g N₂ ? g NH3 = 61.802 cg N₂ x LR 2 mol NH3 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 How much N₂ remains in the vessel? 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 You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. 1 mol H₂ x 2.02 g H₂ 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 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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