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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.

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter10: Solutions
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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,
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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