ll Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant :-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass=138-21 gm/mat. And we make 20 m² of each. Solution :- Reactant-1- Potassium carbonate Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = _20_ _ L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (e) x Volume (V) = 0.200 m³¹ x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Mole x Molar mass = 0.004 molx 147.01 gm² = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n ⇒ n = cx V = 0·220 mel x0·02 L = 0·0044 mol. Molay mass = 138.21 gm/mol. NOW, mol Mass = Molex Molar mass = 0·0044 mdx138.21 gm² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. ? Step2 b) Do 8
ll Freedom 10:31 PM We have to find the mass used in each reactant to create stock solution. Reactant :-1 Potassium carbonate - [0.200M] Molar mass 147.01 gm/mol Reactant-2 calcium chloride dihydrate= [0.220 M] Molay mass=138-21 gm/mat. And we make 20 m² of each. Solution :- Reactant-1- Potassium carbonate Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = _20_ _ L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (e) = Volume (V) Moles (n) = Concentration (e) x Volume (V) = 0.200 m³¹ x 0·02 L = 0·004 mol. Molar mass = 147.01 gm/mol. Now. Mass = Mole x Molar mass = 0.004 molx 147.01 gm² = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (v) = 20 mL = 0.02 L Now, c = n ⇒ n = cx V = 0·220 mel x0·02 L = 0·0044 mol. Molay mass = 138.21 gm/mol. NOW, mol Mass = Molex Molar mass = 0·0044 mdx138.21 gm² = 0.608 gm. Hence, 0.608 gm of calcium chloride dihydrate is used. ? Step2 b) Do 8
Chemistry
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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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Please help using the information below calculate the limiting reagent and the theoretical yield
![Freedom
10:31 PM
We have to find the mass used in each reactant to create stock solution.
Reactant:-1
Potassium carbonate - [0.200M]
Molar mass 147.01 gm/mol
Reactant-2
calcium chloride dihydrate= [0.220 M]
Molay mass = 138.21 gm/mol.
And we make 20 m² of each.
Step2
b)
Solution :-
Reactant-1 Potassiuma casi bonate
Concentration (c) = 0·200 M = 0·200 mol/L
Volume (V) = 20 mL = 20 L = 0·02 L [1L = 1000 m²]
1000
Now,
Moles(n)
Concentration (e) =
Volume (V)
Moles (n) = Concentration (C) x Volume (V) = 0-200 mol x 0·02 L = 0·004 mol.
Molar mass = 147.01 gm/mol.
Now.
Mass = Molex Molar mass = 0.004 molx 147.01 gm = 0.588 gm.
Hence, 0.588 gm of potassium carbonate is used.
Reactant-2- Calcium chloride dihydrate
Concentration (e) = 0·220 M = 0·220 mol/L
Volume (v) = 20 mL = 0.02 L
Now,
c = n
n = cx V = 0·220 mol x0·02 L = 0.0044 mol.
Molar mass = 138.21 g/mol.
NOW,
mol
Mass = Molex Molar mass = 0.0044 mdx138.21 m² = 0.608 gm.
Hence, 0.608 gm of calcium chloride dihydrate is used.
√x
Do
8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2488c527-aba7-4d28-abdc-9ddf115764f0%2Fd0f55280-5425-49ab-941f-e9c83a8bb5ea%2Fjjykomfi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Freedom
10:31 PM
We have to find the mass used in each reactant to create stock solution.
Reactant:-1
Potassium carbonate - [0.200M]
Molar mass 147.01 gm/mol
Reactant-2
calcium chloride dihydrate= [0.220 M]
Molay mass = 138.21 gm/mol.
And we make 20 m² of each.
Step2
b)
Solution :-
Reactant-1 Potassiuma casi bonate
Concentration (c) = 0·200 M = 0·200 mol/L
Volume (V) = 20 mL = 20 L = 0·02 L [1L = 1000 m²]
1000
Now,
Moles(n)
Concentration (e) =
Volume (V)
Moles (n) = Concentration (C) x Volume (V) = 0-200 mol x 0·02 L = 0·004 mol.
Molar mass = 147.01 gm/mol.
Now.
Mass = Molex Molar mass = 0.004 molx 147.01 gm = 0.588 gm.
Hence, 0.588 gm of potassium carbonate is used.
Reactant-2- Calcium chloride dihydrate
Concentration (e) = 0·220 M = 0·220 mol/L
Volume (v) = 20 mL = 0.02 L
Now,
c = n
n = cx V = 0·220 mol x0·02 L = 0.0044 mol.
Molar mass = 138.21 g/mol.
NOW,
mol
Mass = Molex Molar mass = 0.0044 mdx138.21 m² = 0.608 gm.
Hence, 0.608 gm of calcium chloride dihydrate is used.
√x
Do
8
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