Ca(OH)2 (aq) + H₂SO4 (aq) →+ 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H.SO, 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 H₂0 = 1.839 g Ca(OH)2 x 1 mol Ca(OH)2 x 74.10 g Ca(OH)2 ? g H₂O 25.0 mL H₂SO4 x = Ca (OH)₂ amount = 1.839 g GIVEN 1.469 g 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 = -- 1L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂0 1000 mL H₂SO4 1L H₂SO4 1 mol H₂SO4 1 mol H₂O 2 mol H₂O x 18.02 g H₂O = 1 mol Ca(OH)2 1 mol H₂O 1 LH₂ SO4 1000 mL H₂ SOA X LEFT 0.8944 g H₂O given - Ca (OH)2 amount used 0.3705 g USED OVER = EXCESS 0.200 mol H₂ SO4 1 mol Ca (OH)2 1 LH₂ SOA 1 mol H₂ SO₂ X X 0.681 g CaSO4 74.093 g 1 mol Ca (OH)₂ 1 mol Ca (OH)₂
Ca(OH)2 (aq) + H₂SO4 (aq) →+ 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H.SO, 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 H₂0 = 1.839 g Ca(OH)2 x 1 mol Ca(OH)2 x 74.10 g Ca(OH)2 ? g H₂O 25.0 mL H₂SO4 x = Ca (OH)₂ amount = 1.839 g GIVEN 1.469 g 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 = -- 1L H₂SO4 x 0.200 mol H₂SO4 x 2 mol H₂O x 18.02 g H₂O = 0.180 g H₂0 1000 mL H₂SO4 1L H₂SO4 1 mol H₂SO4 1 mol H₂O 2 mol H₂O x 18.02 g H₂O = 1 mol Ca(OH)2 1 mol H₂O 1 LH₂ SO4 1000 mL H₂ SOA X LEFT 0.8944 g H₂O given - Ca (OH)2 amount used 0.3705 g USED OVER = EXCESS 0.200 mol H₂ SO4 1 mol Ca (OH)2 1 LH₂ SOA 1 mol H₂ SO₂ X X 0.681 g CaSO4 74.093 g 1 mol Ca (OH)₂ 1 mol Ca (OH)₂
Chemistry
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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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Question
Fill in the blanks using the image.
Transcribed Image Text:Ca(OH)2 (aq) + H₂SO4 (aq) →+ 2 H₂O (1) + CaSO4 (aq) (This is LIMITING REACTANT: H.SO, 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 H₂0 = 1.839 g Ca(OH)2 x 1 mol Ca(OH)2 x
74.10 g Ca(OH)2
? 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 1L 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
= Ca (OH)₂ amount
= 1.839 g GIVEN
= 1.469 g
--
IL H₂ SO4
1000 mL H₂ SO4
2 mol H₂O x 18.02 g H₂O =
1 mol Ca(OH)2 1 mol H₂O
X
LEFT
0.8944 g H₂O
given - Ca (OH)2 amount used
0.3705 g USED
OVER = EXCESS
0.200 mol H₂ SO4 1 mol Ca (OH)2
1 LH₂ SO₁ 1 mol H₂ SO4
X
X
0.681 g CaSO4
74.093 g 1 mol Ca (OH)₂
1 mol Ca (OH)₂
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
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