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Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of calcium hydroxide Grams of sulfuric acid Grams of water Grams of calcium sulfate

Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of calcium hydroxide Grams of sulfuric acid Grams of water Grams of calcium sulfate

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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Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of calcium hydroxide Grams of sulfuric acid Grams of water Grams of calcium sulfate
Transcribed Image Text:Balanced Chemical Equation Reaction Type: At completion of reactions: Grams of calcium hydroxide Grams of sulfuric acid Grams of water Grams of calcium sulfate
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 1 mol Ca(OH)2 x 2 mol H₂O x 18.02 g H₂O = 0.8944 g H₂O 74.10 g Ca(OH)2 1 mol Ca(OH)2 1 mol H₂O ? g H₂O = 25.0 mL H₂SO4 x 1L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂0 1000 mL H₂SO4 IL 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 IL H₂SO4 x 0.200 mol H₂SO4 x 1 mol CaSO4 x 136.14 g CaSO4 1000 mL H₂SO4 1L H₂SO4 1 mol H₂SO4 1 mol CaSO4 ? g Ca (OH)₂ USED = 25.0 mLH₂ SO4 X- = 0.3705 g Ca (OH)2 Amount of Ca(OH)2 remaining in the container —— IL H₂ SOA 1000 mL H₂ SO4 = Ca (OH)₂ amount given = 1.839 g GIVEN - 0.3705 g USED = 1.469 g 0.200 mol H₂ SO4 X- 1 LH₂ SOA - X LEFT OVER = EXCESS 1 mol Ca (OH)₂ 1 mol H₂ SO4 Ca (OH)2 amount used X 0.681 g CaSO4 74.093 g 1 mol Ca (OH)₂2 1 mol Ca (OH)₂
Transcribed Image Text: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 1 mol Ca(OH)2 x 2 mol H₂O x 18.02 g H₂O = 0.8944 g H₂O 74.10 g Ca(OH)2 1 mol Ca(OH)2 1 mol H₂O ? g H₂O = 25.0 mL H₂SO4 x 1L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂0 1000 mL H₂SO4 IL 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 IL H₂SO4 x 0.200 mol H₂SO4 x 1 mol CaSO4 x 136.14 g CaSO4 1000 mL H₂SO4 1L H₂SO4 1 mol H₂SO4 1 mol CaSO4 ? g Ca (OH)₂ USED = 25.0 mLH₂ SO4 X- = 0.3705 g Ca (OH)2 Amount of Ca(OH)2 remaining in the container —— IL H₂ SOA 1000 mL H₂ SO4 = Ca (OH)₂ amount given = 1.839 g GIVEN - 0.3705 g USED = 1.469 g 0.200 mol H₂ SO4 X- 1 LH₂ SOA - X LEFT OVER = EXCESS 1 mol Ca (OH)₂ 1 mol H₂ SO4 Ca (OH)2 amount used X 0.681 g CaSO4 74.093 g 1 mol Ca (OH)₂2 1 mol Ca (OH)₂
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