CaCl,*2H,O(aq) + Na,CO3 (a4) → 2NaCl + CaCO3 () + 2H,0 (aq) 1 mole Cacl,"2H,O 2 mole NacI 58.44g NaCI 5.0 g CaCl,*2H,O x 3.97g Naci 147.02g cacı,"2H,0 1 mole cacı,"2H,0 1 mole Naci 1 mole Na,CO, 58.44g Naci X 2 mole Naci 5.0 g Na,Co, x = 5.51g Naci 105.99 g Na,CO, 1 mole Na,Co, 1 mole Naci 3.97g is the Theoretical Yield of the Precipitate.
CaCl,*2H,O(aq) + Na,CO3 (a4) → 2NaCl + CaCO3 () + 2H,0 (aq) 1 mole Cacl,"2H,O 2 mole NacI 58.44g NaCI 5.0 g CaCl,*2H,O x 3.97g Naci 147.02g cacı,"2H,0 1 mole cacı,"2H,0 1 mole Naci 1 mole Na,CO, 58.44g Naci X 2 mole Naci 5.0 g Na,Co, x = 5.51g Naci 105.99 g Na,CO, 1 mole Na,Co, 1 mole Naci 3.97g is the Theoretical Yield of the Precipitate.
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter3: Chemical Reactions
Section: Chapter Questions
Problem 6PS: A 0.20 mol sample of magnesium burns in air to form 0.20 mol of solid MgO. What amount (moles) of...
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Question
Do you agree with the work above that the Theoretical Yield of the precipitate is 3.97g?
A) Yes, but I would need to know the Actual Yield to know how close we got to the correct answer
B) No, NaCl is not the precipitate so this does not give you an accurate Theoretical Yield of the precipitate
C) No, the molar mass for Calcium Chloride Dihydrate is 110.98g
D) Yes, 3.97 g is the correct Theoretical Yield of the Precipitate
E) No, 5.51 g is the correct Theoretical Yield of the Precipitate
Transcribed Image Text:CaCl,*2H,O(aq) + Na,CO3 (a4) → 2NaCl
+ CaCO3 () + 2H,0
(aq)
1 mole Cacl,"2H,O
2 mole NacI
58.44g NaCI
5.0 g CaCl,*2H,O x
3.97g Naci
147.02g cacı,"2H,0
1 mole cacı,"2H,0
1 mole Naci
1 mole Na,CO,
58.44g Naci
X
2 mole Naci
5.0 g Na,Co, x
= 5.51g Naci
105.99 g Na,CO, 1 mole Na,Co,
1 mole Naci
3.97g is the Theoretical Yield of the Precipitate.
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