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 (2) = Volume (V) => Moles (n) = Concentration (C) x Volume (v) = 0.200 mol x 0.02 L = 0·004 mol. Molar mass = 147.01 g/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 = 꼼 ⇒ n = cx V = 0·220 mal x 0·02 L = 0.0044 mol. Molar mass = 138.21 gm/mol. Now, Mass= Molex Molay mass = 0.0044 mol x 138.21 gm -=0.608 gm. mol Hence, 0.608 gm of calcium chloride dishydrate is used.
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 (2) = Volume (V) => Moles (n) = Concentration (C) x Volume (v) = 0.200 mol x 0.02 L = 0·004 mol. Molar mass = 147.01 g/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 = 꼼 ⇒ n = cx V = 0·220 mal x 0·02 L = 0.0044 mol. Molar mass = 138.21 gm/mol. Now, Mass= Molex Molay mass = 0.0044 mol x 138.21 gm -=0.608 gm. mol Hence, 0.608 gm of calcium chloride dishydrate is used.
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter11: Solutions
Section: Chapter Questions
Problem 50P
Related questions
Question
Please help
Please find the limiting reagent using the information in the photo
Please show your work using
The name of the equation are supposed to be switched so please take not of that
![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 (2) =
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,
☆
⇒ n = cx V = 0·220 mol x 0·02 L = 0.0044 mol.
Molar mass = 138.21 gm/mol.
Mass= Molex Molay mass = 0.0044 mol x 138.21 gm =
-=0.608 gm.
mol
Hence, 0.608 gm of calcium chloride dishydrate is used.
Now,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F826a5dfc-503b-435a-b61c-90a3cd1f56dd%2F31f4e9dc-8f28-42d9-8c09-05beeee11f47%2Flni92xd_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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 (2) =
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,
☆
⇒ n = cx V = 0·220 mol x 0·02 L = 0.0044 mol.
Molar mass = 138.21 gm/mol.
Mass= Molex Molay mass = 0.0044 mol x 138.21 gm =
-=0.608 gm.
mol
Hence, 0.608 gm of calcium chloride dishydrate is used.
Now,
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